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Construct a Perfect Binary Tree from Preorder Traversal

  • Last Updated : 22 Jul, 2021

Given an array pre[], representing the Preorder traversal of a Perfect Binary Tree consisting of N nodes, the task is to construct a Perfect Binary Tree from the given Preorder Traversal and return the root of the tree.

Examples:

Input: pre[] = {1, 2, 4, 5, 3, 6, 7}
Output:
                            1
                         /    \
                      /        \
                     2          3
                  /   \       /   \ 
                /      \    /      \
              4       5   6       7

Input: pre[] = {1, 2, 3}
Output:
                            1
                         /    \
                      /        \
                    2          3

Generally to construct a binary tree, we can not do it by only using the preorder traversal, but here an extra condition is given that the binary tree is Perfect binary tree. We can use that extra condition.



For Perfect binary tree every node has either 2 or 0 children , and all the leaf nodes are present at same level. And the preorder traversal of a binary tree contains the root first, then the preorder traversal of the left subtree, then the preorder traversal of the right subtree . So for Perfect binary tree root should have same numbers of children in both subtrees , so the number ( say, n) of elements after the root in the preorder traversal should be even (2 * number of nodes in one subtree , as it is Perfect binary tree) . And since each subtrees have equal number of nodes for Perfect  binary tree, we can find the preorder traversal of the left subtree (which is half of the array after the root in preorder traversal of the whole tree), and we know the preorder traversal of the right subtree must be after the preorder traversal of the left subtree, so rest half is the preorder traversal of the right subtree.

So the first element in the preorder traversal is the root, we will build a node as the root with this element , then we can easily find the preorder traversals of the left and right subtrees of the root, and we will recursively build the left and right subtrees of the root.

 Approach: The given problem can be solved using recursion. Follow the steps below to solve the problem:

  • Create a function, say BuildPerfectBT_helper with parameters as preStart, preEnd, pre[] where preStart represent starting index of the array pre[] and preEnd represents the ending index of the array pre[] and perform the following steps:
    • If the value of preStart is greater than preEnd, then return NULL.
    • Initialize root as pre[preStart].
    • If the value of preStart is the same as the preEnd, then return root.
    • Initialize 4 variable, say leftPreStart as preStart + 1, rightPreStart as leftPreStart + (preEnd – leftPreStart+1)/2, leftPreEnd as rightPreStart – 1 and rightPreEnd as preEnd.
    • Modify the value of root->left by recursively calling the function buildPerfectBT_helper() with the parameters leftPreStart, leftPreEnd and pre[].
    • Modify the value of root->right by recursively calling the function buildPerfectBT_helper() with the parameters rightPreStart, rightPreEnd and pre[].
    • After performing the above steps, return root.
  • After creating the Perfect Binary Tree, print the Inorder traversal of the tree.

Below is the illustration of the above steps discussed:

Step 1: build([1, 2, 4, 5, 3, 6, 7])

Step 2:
                     1
                /        \
             /            \
build([2, 4, 5])      build([3, 6, 7])

Now first element (1 here) is root, then the subarray after the first element  ( which is [2,4,5,3,6,7] here ) contains the preorder traversals of the left and right subtrees. And we know left subtree’s preorder traversal is first half , i.e, [2,4,5]  , and the right subtree’s preorder traversal is the second half , i.e, [3,6,7]. Now recursively build the left and right subtrees. 

Step 3:
                                                    1



                                   ___________|_________________

                              /                                               \

                         /                                                      \

                      2                                                          3

           ______/___________                                   _______\____________

        /                            \                               /                               \

    /                                 \                         /                                     \

build([4])                   build([5])            build([6])                          build([7]) 
 
Step 4: 
                   1
                /   \
             /       \
          2         3
        /   \     /   \ 
      /     \   /     \
    4       5  6     7
 

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Structure of the tree
struct Node {
    int data;
    Node *left, *right;
 
    Node(int val)
    {
        data = val;
        left = right = NULL;
    }
};
 
// Function to create a new node with
// the value val
Node* getNewNode(int val)
{
    Node* newNode = new Node(val);
    newNode->data = val;
    newNode->left = newNode->right = NULL;
 
    // Return the newly created node
    return newNode;
}
 
// Function to create the Perfect
// Binary Tree
Node* buildPerfectBT_helper(int preStart,
                            int preEnd,
                            int pre[])
{
    // If preStart > preEnd return NULL
    if (preStart > preEnd)
        return NULL;
 
    // Initialize root as pre[preStart]
    Node* root = getNewNode(pre[preStart]);
    ;
 
    // If the only node is left,
    // then return node
    if (preStart == preEnd)
        return root;
 
    // Parameters for further recursion
    int leftPreStart = preStart + 1;
    int rightPreStart = leftPreStart
                        + (preEnd - leftPreStart + 1) / 2;
    int leftPreEnd = rightPreStart - 1;
    int rightPreEnd = preEnd;
 
    // Recursive Call to build the
    // subtree of root node
    root->left = buildPerfectBT_helper(
        leftPreStart, leftPreEnd, pre);
 
    root->right = buildPerfectBT_helper(
        rightPreStart, rightPreEnd, pre);
 
    // Return the created root
    return root;
}
 
// Function to build Perfect Binary Tree
Node* buildPerfectBT(int pre[], int size)
{
    return buildPerfectBT_helper(0, size - 1, pre);
}
 
// Function to print the Inorder of
// the given Tree
void printInorder(Node* root)
{
    // Base Case
    if (!root)
        return;
 
    // Left Recursive Call
    printInorder(root->left);
 
    // Print the data
    cout << root->data << " ";
 
    // Right Recursive Call
    printInorder(root->right);
}
 
// Driver Code
int main()
{
    int pre[] = { 1, 2, 4, 5, 3, 6, 7 };
    int N = sizeof(pre) / sizeof(pre[0]);
 
    // Function Call
    Node* root = buildPerfectBT(pre, N);
 
    // Print Inorder Traversal
    cout << "\nInorder traversal of the tree: ";
    printInorder(root);
 
    return 0;
}
Output: 
Inorder traversal of the tree: 4 2 5 1 6 3 7

 

Time Complexity: O(N)
Auxiliary Space: O(N)

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