Construct a Binary Tree from Postorder and Inorder

Given Postorder and Inorder traversals, construct the tree.

Examples:

Input: 
in[]   = {2, 1, 3}
post[] = {2, 3, 1}

Output: Root of below tree

      1
    /   \
  2     3 

Input:
in[]   = {4, 8, 2, 5, 1, 6, 3, 7}
post[] = {8, 4, 5, 2, 6, 7, 3, 1} 

Output: Root of below tree

           1
        /     \
     2        3
  /    \    /   \
4     5   6    7
  \
   8

Approach: To solve the problem follow the below idea:

Note: We have already discussed the construction of trees from Inorder and Preorder traversals:

Follow the below steps:

Let us see the process of constructing a tree from in[] = {4, 8, 2, 5, 1, 6, 3, 7} and post[] = {8, 4, 5, 2, 6, 7, 3, 1}:

• We first find the last node in post[]. The last node is “1”, we know this value is the root as the root always appears at the end of postorder traversal.
• We search “1” in in[] to find the left and right subtrees of the root. Everything on the left of “1” in in[] is in the left subtree and everything on right is in the right subtree. 

            1
          /    \
[4, 8, 2, 5]   [6, 3, 7]

• We recur the above process for the following two. 
  • Recur for in[] = {6, 3, 7} and post[] = {6, 7, 3} Make the created tree as right child of root. 
  • Recur for in[] = {4, 8, 2, 5} and post[] = {8, 4, 5, 2}. Make the created tree the left child of the root.

Note: One important observation is, that we recursively call for the right subtree before the left subtree as we decrease the index of the postorder index whenever we create a new node

Below is the implementation of the above approach: 

C++

/* C++ program to construct tree using inorder and
   postorder traversals */
#include <bits/stdc++.h>

using namespace std;

/* A binary tree node has data, pointer to left
   child and a pointer to right child */
struct Node {
    int data;
    Node *left, *right;
};

// Utility function to create a new node
Node* newNode(int data);

/* Prototypes for utility functions */
int search(int arr[], int strt, int end, int value);

/* Recursive function to construct binary of size n
   from  Inorder traversal in[] and Postorder traversal
   post[].  Initial values of inStrt and inEnd should
   be 0 and n -1.  The function doesn't do any error
   checking for cases where inorder and postorder
   do not form a tree */
Node* buildUtil(int in[], int post[], int inStrt, int inEnd,
                int* pIndex)
{
    // Base case
    if (inStrt > inEnd)
        return NULL;

    /* Pick current node from Postorder traversal using
       postIndex and decrement postIndex */
    Node* node = newNode(post[*pIndex]);
    (*pIndex)--;

    /* If this node has no children then return */
    if (inStrt == inEnd)
        return node;

    /* Else find the index of this node in Inorder
       traversal */
    int iIndex = search(in, inStrt, inEnd, node->data);

    /* Using index in Inorder traversal, construct left and
       right subtrees */
    node->right
        = buildUtil(in, post, iIndex + 1, inEnd, pIndex);
    node->left
        = buildUtil(in, post, inStrt, iIndex - 1, pIndex);

    return node;
}

// This function mainly initializes index of root
// and calls buildUtil()
Node* buildTree(int in[], int post[], int n)
{
    int pIndex = n - 1;
    return buildUtil(in, post, 0, n - 1, &pIndex);
}

/* Function to find index of value in arr[start...end]
   The function assumes that value is postsent in in[] */
int search(int arr[], int strt, int end, int value)
{
    int i;
    for (i = strt; i <= end; i++) {
        if (arr[i] == value)
            break;
    }
    return i;
}

/* Helper function that allocates a new node */
Node* newNode(int data)
{
    Node* node = (Node*)malloc(sizeof(Node));
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}

/* This function is here just to test  */
void preOrder(Node* node)
{
    if (node == NULL)
        return;
    printf("%d ", node->data);
    preOrder(node->left);
    preOrder(node->right);
}

// Driver code
int main()
{
    int in[] = { 4, 8, 2, 5, 1, 6, 3, 7 };
    int post[] = { 8, 4, 5, 2, 6, 7, 3, 1 };
    int n = sizeof(in) / sizeof(in[0]);

    Node* root = buildTree(in, post, n);

    cout << "Preorder of the constructed tree : \n";
    preOrder(root);

    return 0;
}

Java

// Java program to construct a tree using inorder
// and postorder traversals

/* A binary tree node has data, pointer to left
   child and a pointer to right child */
class Node {
    int data;
    Node left, right;

    public Node(int data)
    {
        this.data = data;
        left = right = null;
    }
}

class BinaryTree {
    /* Recursive function to construct binary of size n
       from  Inorder traversal in[] and Postorder traversal
       post[]. Initial values of inStrt and inEnd should
       be 0 and n -1.  The function doesn't do any error
       checking for cases where inorder and postorder
       do not form a tree */
    Node buildUtil(int in[], int post[], int inStrt,
                   int inEnd, int postStrt, int postEnd)
    {
        // Base case
        if (inStrt > inEnd)
            return null;

        /* Pick current node from Postorder traversal using
           postIndex and decrement postIndex */
        Node node = new Node(post[postEnd]);

        /* If this node has no children then return */
        if (inStrt == inEnd)
            return node;
        int iIndex = search(in, inStrt, inEnd, node.data);

        /* Using index in Inorder traversal, construct left
           and right subtrees */
        node.left = buildUtil(
            in, post, inStrt, iIndex - 1, postStrt,
            postStrt - inStrt + iIndex - 1);
        node.right = buildUtil(in, post, iIndex + 1, inEnd,
                               postEnd - inEnd + iIndex,
                               postEnd - 1);

        return node;
    }

    /* Function to find index of value in arr[start...end]
       The function assumes that value is postsent in in[]
     */
    int search(int arr[], int strt, int end, int value)
    {
        int i;
        for (i = strt; i <= end; i++) {
            if (arr[i] == value)
                break;
        }
        return i;
    }

    /* This function is here just to test  */
    void preOrder(Node node)
    {
        if (node == null)
            return;
        System.out.print(node.data + " ");
        preOrder(node.left);
        preOrder(node.right);
    }

    // Driver Code
    public static void main(String[] args)
    {
        BinaryTree tree = new BinaryTree();
        int in[] = new int[] { 4, 8, 2, 5, 1, 6, 3, 7 };
        int post[] = new int[] { 8, 4, 5, 2, 6, 7, 3, 1 };
        int n = in.length;
        Node root
            = tree.buildUtil(in, post, 0, n - 1, 0, n - 1);
        System.out.println(
            "Preorder of the constructed tree : ");
        tree.preOrder(root);
    }
}

// This code has been contributed by Mayank
// Jaiswal(mayank_24)

Python3

# Python3 program to construct tree using
# inorder and postorder traversals

# Helper function that allocates
# a new node


class newNode:
    def __init__(self, data):
        self.data = data
        self.left = self.right = None

# Recursive function to construct binary
# of size n from Inorder traversal in[]
# and Postorder traversal post[]. Initial
# values of inStrt and inEnd should be
# 0 and n -1. The function doesn't do any
# error checking for cases where inorder
# and postorder do not form a tree


def buildUtil(In, post, inStrt, inEnd, pIndex):

    # Base case
    if (inStrt > inEnd):
        return None

    # Pick current node from Postorder traversal
    # using postIndex and decrement postIndex
    node = newNode(post[pIndex[0]])
    pIndex[0] -= 1

    # If this node has no children
    # then return
    if (inStrt == inEnd):
        return node

    # Else find the index of this node
    # in Inorder traversal
    iIndex = search(In, inStrt, inEnd, node.data)

    # Using index in Inorder traversal,
    # construct left and right subtress
    node.right = buildUtil(In, post, iIndex + 1,
                           inEnd, pIndex)
    node.left = buildUtil(In, post, inStrt,
                          iIndex - 1, pIndex)

    return node

# This function mainly initializes index
# of root and calls buildUtil()


def buildTree(In, post, n):
    pIndex = [n - 1]
    return buildUtil(In, post, 0, n - 1, pIndex)

# Function to find index of value in
# arr[start...end]. The function assumes
# that value is postsent in in[]


def search(arr, strt, end, value):
    i = 0
    for i in range(strt, end + 1):
        if (arr[i] == value):
            break
    return i

# This function is here just to test


def preOrder(node):
    if node == None:
        return
    print(node.data, end=" ")
    preOrder(node.left)
    preOrder(node.right)


# Driver code
if __name__ == '__main__':
    In = [4, 8, 2, 5, 1, 6, 3, 7]
    post = [8, 4, 5, 2, 6, 7, 3, 1]
    n = len(In)

    root = buildTree(In, post, n)

    print("Preorder of the constructed tree :")
    preOrder(root)

# This code is contributed by PranchalK

C#

// C# program to construct a tree using
// inorder and postorder traversals
using System;

/* A binary tree node has data, pointer
to left child and a pointer to right child */
public class Node {
    public int data;
    public Node left, right;

    public Node(int data)
    {
        this.data = data;
        left = right = null;
    }
}

// Class Index created to implement
// pass by reference of Index
public class Index {
    public int index;
}

class GFG {
    /* Recursive function to construct binary
    of size n from Inorder traversal in[] and
    Postorder traversal post[]. Initial values
    of inStrt and inEnd should be 0 and n -1.
    The function doesn't do any error checking
    for cases where inorder and postorder do
    not form a tree */
    public virtual Node buildUtil(int[] @in, int[] post,
                                  int inStrt, int inEnd,
                                  Index pIndex)
    {
        // Base case
        if (inStrt > inEnd) {
            return null;
        }

        /* Pick current node from Postorder traversal
        using postIndex and decrement postIndex */
        Node node = new Node(post[pIndex.index]);
        (pIndex.index)--;

        /* If this node has no children
        then return */
        if (inStrt == inEnd) {
            return node;
        }

        /* Else find the index of this node
        in Inorder traversal */
        int iIndex = search(@in, inStrt, inEnd, node.data);

        /* Using index in Inorder traversal,
        construct left and right subtrees */
        node.right = buildUtil(@in, post, iIndex + 1, inEnd,
                               pIndex);
        node.left = buildUtil(@in, post, inStrt, iIndex - 1,
                              pIndex);

        return node;
    }

    // This function mainly initializes
    // index of root and calls buildUtil()
    public virtual Node buildTree(int[] @in, int[] post,
                                  int n)
    {
        Index pIndex = new Index();
        pIndex.index = n - 1;
        return buildUtil(@in, post, 0, n - 1, pIndex);
    }

    /* Function to find index of value in
    arr[start...end]. The function assumes
    that value is postsent in in[] */
    public virtual int search(int[] arr, int strt, int end,
                              int value)
    {
        int i;
        for (i = strt; i <= end; i++) {
            if (arr[i] == value) {
                break;
            }
        }
        return i;
    }

    /* This function is here just to test */
    public virtual void preOrder(Node node)
    {
        if (node == null) {
            return;
        }
        Console.Write(node.data + " ");
        preOrder(node.left);
        preOrder(node.right);
    }

    // Driver Code
    public static void Main(string[] args)
    {
        GFG tree = new GFG();
        int[] @in = new int[] { 4, 8, 2, 5, 1, 6, 3, 7 };
        int[] post = new int[] { 8, 4, 5, 2, 6, 7, 3, 1 };
        int n = @in.Length;
        Node root = tree.buildTree(@in, post, n);
        Console.WriteLine(
            "Preorder of the constructed tree : ");
        tree.preOrder(root);
    }
}

// This code is contributed by Shrikant13

Javascript

<script>
// Javascript program to construct a tree using inorder
// and postorder traversals

    /* A binary tree node has data, pointer to left
   child and a pointer to right child */
    class Node
    {
        constructor(data)
        {
            this.data = data;
            this.left = this.right = null;
        }
    }

    /* Recursive function to construct binary of size n
       from  Inorder traversal in[] and Postorder traversal
       post[]. Initial values of inStrt and inEnd should
       be 0 and n -1.  The function doesn't do any error
       checking for cases where inorder and postorder
       do not form a tree */
    function buildUtil(In, post, inStrt, inEnd, postStrt, postEnd)
    {

        // Base case
        if (inStrt > inEnd)
            return null;

        /* Pick current node from Postorder traversal using
           postIndex and decrement postIndex */
        let node = new Node(post[postEnd]);

        /* If this node has no children then return */
        if (inStrt == inEnd)
            return node;
        let iIndex = search(In, inStrt, inEnd, node.data);

        /* Using index in Inorder traversal, construct left
           and right subtrees */
        node.left = buildUtil(
            In, post, inStrt, iIndex - 1, postStrt,
            postStrt - inStrt + iIndex - 1);
        node.right = buildUtil(In, post, iIndex + 1, inEnd,
                               postEnd - inEnd + iIndex,
                               postEnd - 1);

        return node;
    }

    /* Function to find index of value in arr[start...end]
       The function assumes that value is postsent in in[]
     */
    function search(arr,strt,end,value)
    {
        let i;
        for (i = strt; i <= end; i++) {
            if (arr[i] == value)
                break;
        }
        return i;
    }

    /* This function is here just to test  */
    function preOrder(node)
    {
        if (node == null)
            return;
        document.write(node.data + " ");
        preOrder(node.left);
        preOrder(node.right);
    }

    // Driver Code
    let In=[4, 8, 2, 5, 1, 6, 3, 7];
    let post=[8, 4, 5, 2, 6, 7, 3, 1];
    let n = In.length;
    let root
            = buildUtil(In, post, 0, n - 1, 0, n - 1);
    document.write(
            "Preorder of the constructed tree : <br>");
    preOrder(root);

    // This code is contributed by unknown2108
</script>

Output

Preorder of the constructed tree : 
1 2 4 8 5 3 6 7

Time Complexity: O(N²), Where N is the length of the given inorder array
Auxiliary Space: O(N), for recursive call stack

Construct a Binary Tree from Postorder and Inorder using hashing:

To solve the problem follow the below idea:

We can optimize the above solution using hashing. We store indexes of inorder traversal in a hash table. So that search can be done O(1) time If given that element in the tree is not repeated.

Follow the below steps to solve the problem:

• We first find the last node in post[]. The last node is “1”, we know this value is the root as the root always appears at the end of postorder traversal.
• we get the index of postorder[i], in inorder using the map to find the left and right subtrees of the root. Everything on the left of “1” in in[] is in the left subtree and everything on right is in the right subtree. 
• We recur the above process for the following two. 
  • Recur for in[] = {6, 3, 7} and post[] = {6, 7, 3} Make the created tree as right child of root. 
  • Recur for in[] = {4, 8, 2, 5} and post[] = {8, 4, 5, 2}. Make the created tree the left child of the root.

Below is the implementation of the above approach:

C++

/* C++ program to construct tree using inorder and
postorder traversals */
#include <bits/stdc++.h>

using namespace std;

/* A binary tree node has data, pointer to left
child and a pointer to right child */
struct Node {
    int data;
    Node *left, *right;
};

// Utility function to create a new node
Node* newNode(int data);

/* Recursive function to construct binary of size n
from Inorder traversal in[] and Postorder traversal
post[]. Initial values of inStrt and inEnd should
be 0 and n -1. The function doesn't do any error
checking for cases where inorder and postorder
do not form a tree */
Node* buildUtil(int in[], int post[], int inStrt, int inEnd,
                int* pIndex, unordered_map<int, int>& mp)
{
    // Base case
    if (inStrt > inEnd)
        return NULL;

    /* Pick current node from Postorder traversal
    using postIndex and decrement postIndex */
    int curr = post[*pIndex];
    Node* node = newNode(curr);
    (*pIndex)--;

    /* If this node has no children then return */
    if (inStrt == inEnd)
        return node;

    /* Else find the index of this node in Inorder
    traversal */
    int iIndex = mp[curr];

    /* Using index in Inorder traversal, construct
    left and right subtrees */
    node->right = buildUtil(in, post, iIndex + 1, inEnd,
                            pIndex, mp);
    node->left = buildUtil(in, post, inStrt, iIndex - 1,
                           pIndex, mp);

    return node;
}

// This function mainly creates an unordered_map, then
// calls buildTreeUtil()
struct Node* buildTree(int in[], int post[], int len)
{
    // Store indexes of all items so that we
    // we can quickly find later
    unordered_map<int, int> mp;
    for (int i = 0; i < len; i++)
        mp[in[i]] = i;

    int index = len - 1; // Index in postorder
    return buildUtil(in, post, 0, len - 1, &index, mp);
}

/* Helper function that allocates a new node */
Node* newNode(int data)
{
    Node* node = (Node*)malloc(sizeof(Node));
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}

/* This function is here just to test */
void preOrder(Node* node)
{
    if (node == NULL)
        return;
    printf("%d ", node->data);
    preOrder(node->left);
    preOrder(node->right);
}

// Driver code
int main()
{
    int in[] = { 4, 8, 2, 5, 1, 6, 3, 7 };
    int post[] = { 8, 4, 5, 2, 6, 7, 3, 1 };
    int n = sizeof(in) / sizeof(in[0]);

    Node* root = buildTree(in, post, n);

    cout << "Preorder of the constructed tree : \n";
    preOrder(root);

    return 0;
}

Java

/* Java program to construct tree using inorder and
postorder traversals */
import java.util.*;
class GFG {

    /* A binary tree node has data, pointer to left
    child and a pointer to right child */
    static class Node {
        int data;
        Node left, right;
    };

    // Utility function to create a new node
    /* Helper function that allocates a new node */
    static Node newNode(int data)
    {
        Node node = new Node();
        node.data = data;
        node.left = node.right = null;
        return (node);
    }

    /* Recursive function to construct binary of size n
    from Inorder traversal in[] and Postorder traversal
    post[]. Initial values of inStrt and inEnd should
    be 0 and n -1. The function doesn't do any error
    checking for cases where inorder and postorder
    do not form a tree */
    static Node buildUtil(int in[], int post[], int inStrt,
                          int inEnd)
    {

        // Base case
        if (inStrt > inEnd)
            return null;

        /* Pick current node from Postorder traversal
        using postIndex and decrement postIndex */
        int curr = post[index];
        Node node = newNode(curr);
        (index)--;

        /* If this node has no children then return */
        if (inStrt == inEnd)
            return node;

        /* Else find the index of this node in Inorder
        traversal */
        int iIndex = mp.get(curr);

        /* Using index in Inorder traversal, con
        left and right subtrees */
        node.right = buildUtil(in, post, iIndex + 1, inEnd);
        node.left = buildUtil(in, post, inStrt, iIndex - 1);
        return node;
    }
    static HashMap<Integer, Integer> mp
        = new HashMap<Integer, Integer>();
    static int index;

    // This function mainly creates an unordered_map, then
    // calls buildTreeUtil()
    static Node buildTree(int in[], int post[], int len)
    {

        // Store indexes of all items so that we
        // we can quickly find later
        for (int i = 0; i < len; i++)
            mp.put(in[i], i);

        index = len - 1; // Index in postorder
        return buildUtil(in, post, 0, len - 1);
    }

    /* This function is here just to test */
    static void preOrder(Node node)
    {
        if (node == null)
            return;
        System.out.printf("%d ", node.data);
        preOrder(node.left);
        preOrder(node.right);
    }

    // Driver code
    public static void main(String[] args)
    {
        int in[] = { 4, 8, 2, 5, 1, 6, 3, 7 };
        int post[] = { 8, 4, 5, 2, 6, 7, 3, 1 };
        int n = in.length;
        Node root = buildTree(in, post, n);
        System.out.print(
            "Preorder of the constructed tree : \n");
        preOrder(root);
    }
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program to construct tree using inorder
# and postorder traversals

# A binary tree node has data, pointer to left
# child and a pointer to right child


class Node:

    def __init__(self, x):

        self.data = x
        self.left = None
        self.right = None

# Recursive function to construct binary of size n
# from Inorder traversal in[] and Postorder traversal
# post[]. Initial values of inStrt and inEnd should
# be 0 and n -1. The function doesn't do any error
# checking for cases where inorder and postorder
# do not form a tree


def buildUtil(inn, post, innStrt, innEnd):

    global mp, index

    # Base case
    if (innStrt > innEnd):
        return None

    # Pick current node from Postorder traversal
    # using postIndex and decrement postIndex
    curr = post[index]
    node = Node(curr)
    index -= 1

    # If this node has no children then return
    if (innStrt == innEnd):
        return node

    # Else find the index of this node inn
    # Inorder traversal
    iIndex = mp[curr]

    # Using index in Inorder traversal,
    # construct left and right subtrees
    node.right = buildUtil(inn, post,
                           iIndex + 1, innEnd)
    node.left = buildUtil(inn, post, innStrt,
                          iIndex - 1)

    return node

# This function mainly creates an unordered_map,
# then calls buildTreeUtil()


def buildTree(inn, post, lenn):

    global index

    # Store indexes of all items so that we
    # we can quickly find later
    for i in range(lenn):
        mp[inn[i]] = i

    # Index in postorder
    index = lenn - 1
    return buildUtil(inn, post, 0, lenn - 1)

# This function is here just to test


def preOrder(node):

    if (node == None):
        return

    print(node.data, end=" ")
    preOrder(node.left)
    preOrder(node.right)


# Driver Code
if __name__ == '__main__':

    inn = [4, 8, 2, 5, 1, 6, 3, 7]
    post = [8, 4, 5, 2, 6, 7, 3, 1]
    n = len(inn)
    mp, index = {}, 0

    root = buildTree(inn, post, n)

    print("Preorder of the constructed tree :")
    preOrder(root)

# This code is contributed by mohit kumar 29

C#

/* C# program to construct tree using inorder and
postorder traversals */
using System;
using System.Collections.Generic;
class GFG {

    /* A binary tree node has data, pointer to left
  child and a pointer to right child */
    public

        class Node {
        public

            int data;
        public

            Node left,
            right;
    };

    // Utility function to create a new node
    /* Helper function that allocates a new node */
    static Node newNode(int data)
    {
        Node node = new Node();
        node.data = data;
        node.left = node.right = null;
        return (node);
    }

    /* Recursive function to construct binary of size n
  from Inorder traversal in[] and Postorder traversal
  post[]. Initial values of inStrt and inEnd should
  be 0 and n -1. The function doesn't do any error
  checking for cases where inorder and postorder
  do not form a tree */
    static Node buildUtil(int[] init, int[] post,
                          int inStrt, int inEnd)
    {

        // Base case
        if (inStrt > inEnd)
            return null;

        /* Pick current node from Postorder traversal
        using postIndex and decrement postIndex */
        int curr = post[index];
        Node node = newNode(curr);
        (index)--;

        /* If this node has no children then return */
        if (inStrt == inEnd)
            return node;

        /* Else find the index of this node in Inorder
        traversal */
        int iIndex = mp[curr];

        /* Using index in Inorder traversal, con
        left and right subtrees */
        node.right
            = buildUtil(init, post, iIndex + 1, inEnd);
        node.left
            = buildUtil(init, post, inStrt, iIndex - 1);
        return node;
    }
    static Dictionary<int, int> mp
        = new Dictionary<int, int>();
    static int index;

    // This function mainly creates an unordered_map, then
    // calls buildTreeUtil()
    static Node buildTree(int[] init, int[] post, int len)
    {

        // Store indexes of all items so that we
        // we can quickly find later
        for (int i = 0; i < len; i++)
            mp.Add(init[i], i);

        index = len - 1; // Index in postorder
        return buildUtil(init, post, 0, len - 1);
    }

    /* This function is here just to test */
    static void preOrder(Node node)
    {
        if (node == null)
            return;
        Console.Write(node.data + " ");
        preOrder(node.left);
        preOrder(node.right);
    }

    // Driver code
    public static void Main(String[] args)
    {
        int[] init = { 4, 8, 2, 5, 1, 6, 3, 7 };
        int[] post = { 8, 4, 5, 2, 6, 7, 3, 1 };
        int n = init.Length;
        Node root = buildTree(init, post, n);
        Console.Write(
            "Preorder of the constructed tree : \n");
        preOrder(root);
    }
}

// This code is contributed by Rajput-Ji

Javascript

<script>

      /* JavaScript program to construct tree using inorder and
      postorder traversals */
      /* A binary tree node has data, pointer to left
      child and a pointer to right child */
      class Node {
        constructor() {
          this.data = 0;
          this.left = null;
          this.right = null;
        }
      }

      // Utility function to create a new node
      /* Helper function that allocates a new node */
      function newNode(data) {
        var node = new Node();
        node.data = data;
        node.left = node.right = null;
        return node;
      }

      /* Recursive function to construct binary of size n
      from Inorder traversal in[] and Postorder traversal
      post[]. Initial values of inStrt and inEnd should
      be 0 and n -1. The function doesn't do any error
      checking for cases where inorder and postorder
      do not form a tree */
      function buildUtil(init, post, inStrt, inEnd) {
        // Base case
        if (inStrt > inEnd) {
          return null;
        }

        /* Pick current node from Postorder traversal
          using postIndex and decrement postIndex */
        var curr = post[index];
        var node = newNode(curr);
        index--;

        /* If this node has no children then return */
        if (inStrt == inEnd) {
          return node;
        }

        /* Else find the index of this node in Inorder
          traversal */
        var iIndex = mp[curr];

        /* Using index in Inorder traversal, con
          left and right subtrees */
        node.right = buildUtil(init, post, iIndex + 1, inEnd);
        node.left = buildUtil(init, post, inStrt, iIndex - 1);
        return node;
      }
      var mp = {};
      var index;

      // This function mainly creates an unordered_map, then
      // calls buildTreeUtil()
      function buildTree(init, post, len) {
        // Store indexes of all items so that we
        // we can quickly find later
        for (var i = 0; i < len; i++) {
          mp[init[i]] = i;
        }

        index = len - 1; // Index in postorder
        return buildUtil(init, post, 0, len - 1);
      }

      /* This function is here just to test */
      function preOrder(node) {
        if (node == null) {
          return;
        }
        document.write(node.data + " ");
        preOrder(node.left);
        preOrder(node.right);
      }

      // Driver code
      var init = [4, 8, 2, 5, 1, 6, 3, 7];
      var post = [8, 4, 5, 2, 6, 7, 3, 1];
      var n = init.length;
      var root = buildTree(init, post, n);
      document.write("Preorder of the constructed tree : <br>");
      preOrder(root);

</script>

Output

Preorder of the constructed tree : 
1 2 4 8 5 3 6 7

Time Complexity: O(N) 
Auxiliary Space: O(N), The extra space is used due to the recursion call stack and to store the elements in the map.

Construct a Binary Tree from Postorder and Inorder using stack and set:

We can use the stack and set without using recursion.

Follow the below steps to solve the problem:

• Create a stack and a set of type Node* and initialize an integer postIndex with N-1
• Run a for loop with p and i, from n-1 to 0
  • Create a new Node with value as postorder[p] and set it as the root node, if it is the first node of our newly created tree
  • Check if the value of stack top is already present in the set, then remove it from the set and set the left child of stack top equal to the new node and pop out the stack top
  • Push the current node into the stack
  • Perform step numbers 3,4 and 5 while p is greater than or equal to zero and postorder[p] is not equal to inorder[i]
  • Set the new node equal to null and while the stack’s top data is equal to the inorder[i], set the node equal to stack top and pop out the stack top
  • If the node is not null then insert the node into the set and push it into the stack also
• Return root of the newly created tree

Below is the implementation of the above idea:

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

/* A binary tree node has data, pointer
to left   child and a pointer to right
child */
struct Node {
    int data;
    Node *left, *right;
    Node(int x)
    {
        data = x;
        left = right = NULL;
    }
};

/*Tree building function*/
Node* buildTree(int in[], int post[], int n)
{

    // Create Stack of type Node*
    stack<Node*> st;

    // Create Set of type Node*
    set<Node*> s;

    // Initialise postIndex with n - 1
    int postIndex = n - 1;

    // Initialise root with NULL
    Node* root = NULL;

    for (int p = n - 1, i = n - 1; p >= 0;) {

        // Initialise node with NULL
        Node* node = NULL;

        // Run do-while loop
        do {

            // Initialise node with
            // new Node(post[p]);
            node = new Node(post[p]);

            // Check is root is
            // equal to NULL
            if (root == NULL) {
                root = node;
            }

            // If size of set
            // is greater than 0
            if (st.size() > 0) {

                // If st.top() is present in the
                // set s
                if (s.find(st.top()) != s.end()) {
                    s.erase(st.top());
                    st.top()->left = node;
                    st.pop();
                }
                else {
                    st.top()->right = node;
                }
            }

            st.push(node);

        } while (post[p--] != in[i] && p >= 0);

        node = NULL;

        // If the stack is not empty and
        // st.top()->data is equal to in[i]
        while (st.size() > 0 && i >= 0
               && st.top()->data == in[i]) {
            node = st.top();

            // Pop elements from stack
            st.pop();
            i--;
        }

        // if node not equal to NULL
        if (node != NULL) {
            s.insert(node);
            st.push(node);
        }
    }

    // Return root
    return root;
}
/* for print preOrder Traversal */
void preOrder(Node* node)
{
    if (node == NULL)
        return;
    printf("%d ", node->data);
    preOrder(node->left);
    preOrder(node->right);
}

// Driver Code
int main()
{

    int in[] = { 4, 8, 2, 5, 1, 6, 3, 7 };
    int post[] = { 8, 4, 5, 2, 6, 7, 3, 1 };
    int n = sizeof(in) / sizeof(in[0]);

    // Function Call
    Node* root = buildTree(in, post, n);

    cout << "Preorder of the constructed tree : \n";

    // Function Call for preOrder
    preOrder(root);
    return 0;
}

Java

// Java program for above approach
import java.io.*;
import java.util.*;

class GFG {

    // Node class
    static class Node {
        int data;
        Node left, right;

        // Constructor
        Node(int x)
        {
            data = x;
            left = right = null;
        }
    }

    // Tree building function
    static Node buildTree(int in[], int post[], int n)
    {

        // Create Stack of type Node*
        Stack<Node> st = new Stack<>();

        // Create HashSet of type Node*
        HashSet<Node> s = new HashSet<>();

        // Initialise postIndex with n - 1
        int postIndex = n - 1;

        // Initialise root with null
        Node root = null;

        for (int p = n - 1, i = n - 1; p >= 0😉 {

            // Initialise node with NULL
            Node node = null;

            // Run do-while loop
            do {

                // Initialise node with
                // new Node(post[p]);
                node = new Node(post[p]);

                // Check is root is
                // equal to NULL
                if (root == null) {
                    root = node;
                }

                // If size of set
                // is greater than 0
                if (st.size() > 0) {

                    // If st.peek() is present in the
                    // set s
                    if (s.contains(st.peek())) {
                        s.remove(st.peek());
                        st.peek().left = node;
                        st.pop();
                    }
                    else {
                        st.peek().right = node;
                    }
                }
                st.push(node);

            } while (post[p--] != in[i] && p >= 0);

            node = null;

            // If the stack is not empty and
            // st.top().data is equal to in[i]
            while (st.size() > 0 && i >= 0
                   && st.peek().data == in[i]) {
                node = st.peek();

                // Pop elements from stack
                st.pop();
                i--;
            }

            // If node not equal to NULL
            if (node != null) {
                s.add(node);
                st.push(node);
            }
        }

        // Return root
        return root;
    }

    // For print preOrder Traversal
    static void preOrder(Node node)
    {
        if (node == null)
            return;

        System.out.printf("%d ", node.data);
        preOrder(node.left);
        preOrder(node.right);
    }

    // Driver Code
    public static void main(String[] args)
    {
        int in[] = { 4, 8, 2, 5, 1, 6, 3, 7 };
        int post[] = { 8, 4, 5, 2, 6, 7, 3, 1 };
        int n = in.length;

        // Function Call
        Node root = buildTree(in, post, n);

        System.out.print(
            "Preorder of the constructed tree : \n");

        // Function Call for preOrder
        preOrder(root);
    }
}

// This code is contributed by sujitmeshram

Python3

# Python program for above approach

# A binary tree node has data, pointer
# to left   child and a pointer to right
# child


class Node:
    def __init__(self, x):
        self.data = x
        self.left = None
        self.right = None

# Tree building function


def buildTree(inorder, post, n):
    # Create Stack of type Node
    st = []
    # Create Set of type Node
    set = []
    # Initialise postIndex with n - 1
    postIndex = n - 1
    # Initialise root with NULL
    root = None

    p = n-1
    i = n-1

    while p >= 0:
        # Initialise node with NULL
        node = None

        #  Run loop
        while True:

            # initialize new node
            node = Node(post[p])

            # check if root is equal to null
            if root == None:
                root = node

            # If size of set is greater than 0
            if len(st) > 0:

                # If st top is present in the set s
                if st[-1] in set:
                    set.remove(st[-1])
                    st[-1].left = node
                    st.pop()
                else:
                    st[-1].right = node

            st.append(node)

            p -= 1
            if post[p+1] == inorder[i] or p < 0:
                break

        node = None

        # If the stack is not empty and st top data is equal to in[i]
        while len(st) > 0 and i >= 0 and st[-1].data == inorder[i]:
            node = st[-1]
            # Pop elements from stack
            st.pop()
            i -= 1

        # if node not equal to None
        if node != None:
            set.append(node)
            st.append(node)

    # Return root
    return root

#  for print preOrder Traversal


def preOrder(node):
    if node == None:
        return
    print(node.data, end=" ")
    preOrder(node.left)
    preOrder(node.right)


# Driver Code
if __name__ == '__main__':
    inorder = [4, 8, 2, 5, 1, 6, 3, 7]
    post = [8, 4, 5, 2, 6, 7, 3, 1]
    n = len(inorder)

    # Function Call
    root = buildTree(inorder, post, n)

    print("Preorder of the constructed tree :")

    # Function Call for preOrder
    preOrder(root)

# This code is contributed by Tapesh(tapeshdua420)

C#

// C# program for above approach
using System;
using System.Collections.Generic;
class GFG {

    // Node class
    public class Node {
        public int data;
        public Node left, right;

        // Constructor
        public Node(int x)
        {
            data = x;
            left = right = null;
        }
    }

    // Tree building function
    static Node buildTree(int[] init, int[] post, int n)
    {

        // Create Stack of type Node*
        Stack<Node> st = new Stack<Node>();

        // Create HashSet of type Node*
        HashSet<Node> s = new HashSet<Node>();

        // Initialise root with null
        Node root = null;
        for (int p = n - 1, i = n - 1; p >= 0;) {

            // Initialise node with NULL
            Node node = null;

            // Run do-while loop
            do {

                // Initialise node with
                // new Node(post[p]);
                node = new Node(post[p]);

                // Check is root is
                // equal to NULL
                if (root == null) {
                    root = node;
                }

                // If size of set
                // is greater than 0
                if (st.Count > 0) {

                    // If st.Peek() is present in the
                    // set s
                    if (s.Contains(st.Peek())) {
                        s.Remove(st.Peek());
                        st.Peek().left = node;
                        st.Pop();
                    }
                    else {
                        st.Peek().right = node;
                    }
                }
                st.Push(node);

            } while (post[p--] != init[i] && p >= 0);

            node = null;

            // If the stack is not empty and
            // st.top().data is equal to in[i]
            while (st.Count > 0 && i >= 0
                   && st.Peek().data == init[i]) {
                node = st.Peek();

                // Pop elements from stack
                st.Pop();
                i--;
            }

            // If node not equal to NULL
            if (node != null) {
                s.Add(node);
                st.Push(node);
            }
        }

        // Return root
        return root;
    }

    // For print preOrder Traversal
    static void preOrder(Node node)
    {
        if (node == null)
            return;

        Console.Write(node.data + " ");
        preOrder(node.left);
        preOrder(node.right);
    }

    // Driver Code
    public static void Main(String[] args)
    {
        int[] init = { 4, 8, 2, 5, 1, 6, 3, 7 };
        int[] post = { 8, 4, 5, 2, 6, 7, 3, 1 };
        int n = init.Length;

        // Function Call
        Node root = buildTree(init, post, n);
        Console.Write(
            "Preorder of the constructed tree : \n");

        // Function Call for preOrder
        preOrder(root);
    }
}

// This code is contributed by aashish1995

Javascript

// JavaScript program for above approach

// Node class
class Node {
  constructor(data) {
    this.data = data;
    this.left = null;
    this.right = null;
  }
}

// Tree building function
function buildTree(inorder, postorder, n) {
  // Create Stack of type Node
  let st = [];

  // Create Set of type Node
  let s = new Set();

  // Initialize postIndex with n - 1
  let postIndex = n - 1;

  // Initialize root with null
  let root = null;

  for (let p = n - 1, i = n - 1; p >= 0; ) {
    // Initialize node with null
    let node = null;

    do {
      // Initialize node with new Node(post[p]);
      node = new Node(postorder[p]);

      // Check if root is equal to null
      if (root === null) {
        root = node;
      }

      // If size of set is greater than 0
      if (st.length > 0)
      {

        // If st[st.length - 1] is present in the set s
        if (s.has(st[st.length - 1])) {
          s.delete(st[st.length - 1]);
          st[st.length - 1].left = node;
          st.pop();
        } else {
          st[st.length - 1].right = node;
        }
      }
      st.push(node);
    } while (postorder[p--] !== inorder[i] && p >= 0);

    node = null;

    // If the stack is not empty and
    // st[st.length - 1].data is equal to in[i]
    while (
      st.length > 0 &&
      i >= 0 &&
      st[st.length - 1].data === inorder[i]
    ) {
      node = st[st.length - 1];

      // Pop elements from stack
      st.pop();
      i--;
    }

    // If node not equal to null
    if (node !== null) {
      s.add(node);
      st.push(node);
    }
  }

  // Return root
  return root;
}

// For print preOrder Traversal
function preOrder(node) {
  if (node === null) return;

  console.log(node.data + " ");
  preOrder(node.left);
  preOrder(node.right);
}

// Test
const inorder = [4, 8, 2, 5, 1, 6, 3, 7];
const postorder = [8, 4, 5, 2, 6, 7, 3, 1];
const n = inorder.length;

// Function Call
const root = buildTree(inorder, postorder, n);

console.log("Preorder of the constructed tree: ");

// Function Call for preOrder
preOrder(root);

// This code is contributed by lokesh.

Output

Preorder of the constructed tree : 
1 2 4 8 5 3 6 7

Time Complexity: O(N)
Auxiliary Space: O(N), The extra space is used to store the elements in the stack and set.