106. Construct Binary Tree from Inorder and Postorder Traversal

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:

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]

Constraints:

1 <= inorder.length <= 3000
postorder.length == inorder.length
-3000 <= inorder[i], postorder[i] <= 3000
inorder and postorder consist of unique values.
Each value of postorder also appears in inorder.
inorder is guaranteed to be the inorder traversal of the tree.
postorder is guaranteed to be the postorder traversal of the tree.

Intuition

Approach:

Recursively build a Tree stiver video

Links

https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/

Video Links

https://www.youtube.com/watch?v=LgLRTaEMRVc&ab_channel=takeUforward

Approach 1:

C++

class Solution {
public:
    unordered_map<int,int> mp;
    TreeNode* Tree(vector<int>& postorder, int p_start, int p_end, vector<int>& inorder, int in_start, int in_end) {
        if(p_start > p_end or in_start > in_end)
            return nullptr;

        TreeNode* root = new TreeNode(postorder[p_start]);
        int index = mp[postorder[p_start]];
        int diff = in_end - index;

        root->left = Tree(postorder, p_start+diff+1, p_end, inorder, in_start, index-1);
        root->right = Tree(postorder, p_start+1, p_start+diff, inorder, index+1, in_end);

        return root;
    }

    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        for(int i=0; i<inorder.size(); i++)
            mp[inorder[i]] = i;
        
        reverse(postorder.begin(), postorder.end());

        return Tree(postorder, 0, postorder.size()-1, inorder, 0, inorder.size()-1);
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++