Case study: Mock interview - construct binary tree using post order traversal and inorder traversal list

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	/*
	Given inorder and postorder traversal of a tree, construct the binary tree.
	Note:
	You may assume that duplicates do not exist in the tree.
	For example, given
	inorder = [9,3,15,20,7]
	postorder = [9,15,7,20,3]
	Return the following binary tree:
	3
	/ \
	9 20
	/ \
	15 7
	*/

view raw constructBinaryTreeUsingPostorderInorderArrays.cpp hosted with ❤ by GitHub

	#include <iostream>
	using namespace std;
	// To execute C++, please define "int main()"
	int main() {
	auto words = { "Hello, ", "World!", "\n" };
	for (const string& word : words) {
	cout << word;
	}
	return 0;
	}
	TreeNode* constructTreeUtil(vector<int> in, vector<int> post, int inStrt,
	int inEnd, int& idx, unordered_map<int,int>& um){
	if(inStrt > inEnd) return nullptr;
	int curr = post[idx];
	auto node = TreeNode(curr); // root node
	idx--;
	if(inStrt == inEnd) return node;// one node - tree with size one ndoe
	int inIdx = um[curr]; // inorder index
	// node->right? why you put node.right first?
	// 3 -> 20 - > 7 root node -> right -> right ->
	// 20's left 15
	// 3's left
	node->right = constructTreeUtil(in, post, inIdx+1, inEnd, idx, um);
	node->left = constructTreeUtil(in, post, inStrt, inIdx-1, idx, um);
	return node;
	}
	TreeNode* constructTree(vector<int> in, vector<int> post){
	unordered_map<int,int>um;
	for(int i=0; i < in.size(); ++i){
	mp[in[i]] = i;
	}
	int idx = in.size()-1;
	return constructTreeUtil(in, post, 0, in.size()-1, idx, um);
	}

view raw ConstructBinaryTreeUsingPostOrderInOrderTraversal.cpp hosted with ❤ by GitHub

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