Construct Binary Tree from Inorder and Postorder Traversal LeetCode Solution

Last updated on October 5th, 2024 at 03:58 pm

Here, We see Construct Binary Tree from Inorder and Postorder Traversal LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Topics

Ordered Map

Companies

Google

Level of Question

Medium

Construct Binary Tree from Inorder and Postorder Traversal LeetCode Solution

Construct Binary Tree from Inorder and Postorder Traversal LeetCode Solution

Problem Statement

Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of the same tree, construct and return the binary tree.

Example 1:

tree

Input: inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
Output: [3,9,20,null,null,15,7]

Example 2:
Input: inorder = [-1], postorder = [-1]
Output: [-1]

1. Construct Binary Tree from Inorder and Postorder Traversal LeetCode Solution C++

class Solution {
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        unordered_map<int, int> index;
        for (int i = 0; i < inorder.size(); i++) {
            index[inorder[i]] = i;
        }
        return buildTreeHelper(inorder, postorder, 0, inorder.size() - 1, 0, postorder.size() - 1, index);
    }
    TreeNode* buildTreeHelper(vector<int>& inorder, vector<int>& postorder, int inorderStart, int inorderEnd, int postorderStart, int postorderEnd, unordered_map<int, int>& index) {
        if (inorderStart > inorderEnd || postorderStart > postorderEnd) {
            return nullptr;
        }
        int rootVal = postorder[postorderEnd];
        TreeNode* root = new TreeNode(rootVal);
        int inorderRootIndex = index[rootVal];
        int leftSubtreeSize = inorderRootIndex - inorderStart;
        root->left = buildTreeHelper(inorder, postorder, inorderStart, inorderRootIndex - 1, postorderStart, postorderStart + leftSubtreeSize - 1, index);
        root->right = buildTreeHelper(inorder, postorder, inorderRootIndex + 1, inorderEnd, postorderStart + leftSubtreeSize, postorderEnd - 1, index);
        return root;
    }
};

2. Construct Binary Tree from Inorder and Postorder Traversal Leetcode Solution Java

class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        return buildTree(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);
    }
    private TreeNode buildTree(int[] inorder, int inStart, int inEnd, int[] postorder, int postStart, int postEnd) {
        if (inStart > inEnd || postStart > postEnd) {
            return null;
        }
        int rootVal = postorder[postEnd];
        TreeNode root = new TreeNode(rootVal);
        int rootIndex = 0;
        for (int i = inStart; i <= inEnd; i++) {
            if (inorder[i] == rootVal) {
                rootIndex = i;
                break;
            }
        }
        int leftSize = rootIndex - inStart;
        int rightSize = inEnd - rootIndex;
        root.left = buildTree(inorder, inStart, rootIndex - 1, postorder, postStart, postStart + leftSize - 1);
        root.right = buildTree(inorder, rootIndex + 1, inEnd, postorder, postEnd - rightSize, postEnd - 1);  
        return root;
    }
}

3. Construct Binary Tree from Inorder and Postorder Traversal Leetcode Solution JavaScript

var buildTree = function(inorder, postorder) {    
    let hash = {};
    for (let i=0;i<inorder.length;i++) hash[inorder[i]] = i;
    let recur = function(start, end) {
        if (start > end) return null;
        let val = postorder.pop();
        let root = new TreeNode(val);
        root.right = recur(hash[val] + 1, end);
        root.left = recur(start, hash[val] - 1);
        return root;
    }
    return recur(0, inorder.length - 1);  
};

4. Construct Binary Tree from Inorder and Postorder Traversal Leetcode Solution Python

class Solution(object):
    def buildTree(self, inorder, postorder):
        if not inorder:
            return None
        root_val = postorder.pop()
        root = TreeNode(root_val)
        inorder_index = inorder.index(root_val)
        root.right = self.buildTree(inorder[inorder_index+1:], postorder)
        root.left = self.buildTree(inorder[:inorder_index], postorder)
        return root
Scroll to Top