Construct a tree from Inorder and Level order traversals | Set 2

Given inorder and level-order traversals of a Binary Tree, construct the Binary Tree. Following is an example to illustrate the problem.

Examples: 

Input: Two arrays that represent Inorder
       and level order traversals of a 
       Binary Tree
in[]    = {4, 8, 10, 12, 14, 20, 22};
level[] = {20, 8, 22, 4, 12, 10, 14};

Output: Construct the tree represented 
        by the two arrays.
        For the above two arrays, the 
        constructed tree is shown.

We have discussed a solution in below post that works in O(N^3) 
Construct a tree from Inorder and Level order traversals | Set 1

Approach: Following algorithm uses O(N^2) time complexity to solve the above problem using the unordered_set data structure in c++ (basically making a hash-table) to put the values of left subtree of the current root and later and we will check in O(1) complexity to find if the current levelOrder node is part of left subtree or not.
If it is the part of left subtree then add in one lLevel array for left otherwise add it to rLevel array for right subtree.

Below is the implementation with the above idea 

C++

/* program to construct tree using inorder
   and levelorder traversals */
#include <bits/stdc++.h>
using namespace std;

/* A binary tree node */
struct Node
{
    int key;
    struct Node* left, *right;
};

Node* makeNode(int data){
    Node* newNode = new Node();
    newNode->key = data;
    newNode->right = newNode->right = NULL;
    return newNode;
}

// Function to build tree from given
// levelorder and inorder
Node* buildTree(int inorder[], int levelOrder[],
                int iStart, int iEnd, int n)
{
    if (n <= 0)
          return NULL;

    // First node of level order is root
    Node* root = makeNode(levelOrder[0]);

    // Search root in inorder
    int index = -1;
    for (int i=iStart; i<=iEnd; i++){
        if (levelOrder[0] == inorder[i]){
            index = i;
            break;
        }
    }

    // Insert all left nodes in hash table
    unordered_set<int> s;
    for (int i=iStart;i<index;i++)
        s.insert(inorder[i]);

    // Separate level order traversals
    // of left and right subtrees.
    int lLevel[s.size()];  // Left
    int rLevel[iEnd-iStart-s.size()]; // Right
    int li = 0, ri = 0;
    for (int i=1;i<n;i++) {
        if (s.find(levelOrder[i]) != s.end())
            lLevel[li++] = levelOrder[i];
        else
            rLevel[ri++] = levelOrder[i];
    }

    // Recursively build left and right
    // subtrees and return root.
    root->left = buildTree(inorder, lLevel,
                 iStart, index-1, index-iStart);
    root->right = buildTree(inorder, rLevel,
                  index+1, iEnd, iEnd-index);
    return root;

}

/* Utility function to print inorder
traversal of binary tree */
void printInorder(Node* node)
{
    if (node == NULL)
       return;
    printInorder(node->left);
    cout << node->key << " ";
    printInorder(node->right);
}

// Driver Code
int main()
{
    int in[] = {4, 8, 10, 12, 14, 20, 22};
    int level[] = {20, 8, 22, 4, 12, 10, 14};
    int n = sizeof(in)/sizeof(in[0]);
    Node *root = buildTree(in, level, 0,
                           n - 1, n);

    /* Let us test the built tree by
     printing Inorder traversal */
    cout << "Inorder traversal of the "
            "constructed tree is \n";
    printInorder(root);

    return 0;
}

Java

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

  /* A binary tree node */
  static class Node {
    int key;
    Node left, right;
  };

  static Node makeNode(int data) {
    Node newNode = new Node();
    newNode.key = data;
    newNode.right = newNode.right = null;
    return newNode;
  }

  // Function to build tree from given
  // levelorder and inorder
  static Node buildTree(int inorder[], int levelOrder[], int iStart, int iEnd, int n) {
    if (n <= 0)
      return null;

    // First node of level order is root
    Node root = makeNode(levelOrder[0]);

    // Search root in inorder
    int index = -1;
    for (int i = iStart; i <= iEnd; i++) {
      if (levelOrder[0] == inorder[i]) {
        index = i;
        break;
      }
    }

    // Insert all left nodes in hash table
    HashSet<Integer> s = new HashSet<>();
    for (int i = iStart; i < index; i++)
      s.add(inorder[i]);

    // Separate level order traversals
    // of left and right subtrees.
    int[] lLevel = new int[s.size()]; // Left
    int[] rLevel = new int[iEnd - iStart - s.size()]; // Right
    int li = 0, ri = 0;
    for (int i = 1; i < n; i++) {
      if (s.contains(levelOrder[i]))
        lLevel[li++] = levelOrder[i];
      else
        rLevel[ri++] = levelOrder[i];
    }

    // Recursively build left and right
    // subtrees and return root.
    root.left = buildTree(inorder, lLevel, iStart, index - 1, index - iStart);
    root.right = buildTree(inorder, rLevel, index + 1, iEnd, iEnd - index);
    return root;

  }

  /*
     * Utility function to print inorder
     * traversal of binary tree
     */
  static void printInorder(Node node) {
    if (node == null)
      return;
    printInorder(node.left);
    System.out.print(node.key + " ");
    printInorder(node.right);
  }

  // Driver Code
  public static void main(String[] args) {

    int in[] = { 4, 8, 10, 12, 14, 20, 22 };
    int level[] = { 20, 8, 22, 4, 12, 10, 14 };
    int n = in.length;
    Node root = buildTree(in, level, 0, n - 1, n);

    /*
         * Let us test the built tree by
         * printing Inorder traversal
         */
    System.out.println("Inorder traversal of the constructed tree is ");
    printInorder(root);

  }
}

// This code is contributed by sanjeev2552

Javascript

<script>
    /*
     program to construct tree using inorder
     and levelorder traversals
     */

     /* A binary tree node */
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.key = data;
        }
    }

    function makeNode(data) {
      let newNode = new Node(data);
      return newNode;
    }

    // Function to build tree from given
    // levelorder and inorder
    function buildTree(inorder, levelOrder, iStart, iEnd, n) {
      if (n <= 0)
        return null;

      // First node of level order is root
      let root = makeNode(levelOrder[0]);

      // Search root in inorder
      let index = -1;
      for (let i = iStart; i <= iEnd; i++) {
        if (levelOrder[0] == inorder[i]) {
          index = i;
          break;
        }
      }

      // Insert all left nodes in hash table
      let s = new Set();
      for (let i = iStart; i < index; i++)
        s.add(inorder[i]);

      // Separate level order traversals
      // of left and right subtrees.
      let lLevel = new Array(s.size); // Left
      let rLevel = new Array(iEnd - iStart - s.size); // Right
      let li = 0, ri = 0;
      for (let i = 1; i < n; i++) {
        if (s.has(levelOrder[i]))
          lLevel[li++] = levelOrder[i];
        else
          rLevel[ri++] = levelOrder[i];
      }

      // Recursively build left and right
      // subtrees and return root.
      root.left = buildTree(inorder, lLevel, iStart, index - 1,
      index - iStart);
      root.right = buildTree(inorder, rLevel, index + 1, iEnd,
      iEnd - index);
      return root;

    }

    /*
       * Utility function to print inorder
       * traversal of binary tree
       */
    function printInorder(node) {
      if (node == null)
        return;
      printInorder(node.left);
      document.write(node.key + " ");
      printInorder(node.right);
    }

    let In = [ 4, 8, 10, 12, 14, 20, 22 ];
    let level = [ 20, 8, 22, 4, 12, 10, 14 ];
    let n = In.length;
    let root = buildTree(In, level, 0, n - 1, n);

    /*
         * Let us test the built tree by
         * printing Inorder traversal
         */
    document.write("Inorder traversal of the constructed tree is " +
    "</br>");
    printInorder(root);

</script>

Output

Inorder traversal of the constructed tree is 
4 8 10 12 14 20 22 

Time Complexity: O(N^2)

Optimization: Without separate level order traversal arrays of left and right subtrees.

Approach: Using Hashing.

Use a HashMap to store the indices of level order traversal. In the range of start & end from inorder, search the element with the least index from the level order map. Create left and right subtrees recursively.

index -> the least index
for left subtree: start to index-1
for right subtree: index+1 to end

Implementation:

Java

import java.util.HashMap;

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

public class ConstructTree {

    //hashmap to store the indices of levelorder array
    HashMap<Integer,Integer> map = new HashMap<>();

    //function to construct hashmap
    void constructMap(int level[]) {
        for(int i=0;i<level.length;i++)
            map.put(level[i], i);
    }

    //function to construct binary tree from inorder and levelorder
    Node construct(int in[],int start,int end) {
        //if start is greater than end return null
        if(start > end)
            return null;

        int min_index = start;

        //In the range of start & end from inorder, search the element
        //with least index from level order map
        for(int i=start+1;i<=end;i++) {
            int temp = in[i];
            //if current element from inorder have least index in
            //levelorder map, update min_index
            if(map.get(in[min_index]) > map.get(temp))
                min_index = i;
        }

        //create a node with current element
        Node root = new Node(in[min_index]);

        //if start is equal to end, then return root
        if(start == end)
            return root;

        //construct left and right subtrees
        root.left = construct(in,start,min_index-1);
        root.right = construct(in,min_index+1,end);

        return root;
    }

    //function to print inorder
    void printInorder(Node node) {
        if (node == null)
          return;
        printInorder(node.left);
        System.out.print(node.data + " ");
        printInorder(node.right);
      }

    //Driver function
    public static void main(String[] args) {
        ConstructTree tree = new ConstructTree();

        int in[]    = {4, 8, 10, 12, 14, 20, 22};
        int level[] = {20, 8, 22, 4, 12, 10, 14};

        //function calls
        tree.constructMap(level);
        int n = level.length;
        Node root = tree.construct(in, 0, n-1);
        tree.printInorder(root);
    }
}

//This method is contributed by Likhita AVL

Output

4 8 10 12 14 20 22 

Time Complexity: O(n^2).