---
tags: leetcode
---

# [889. Construct Binary Tree from Preorder and Postorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-postorder-traversal/)
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# My Solution
## The Key Idea for Solving This Coding Question
## C++ Code

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */

class Solution {
public:
    TreeNode *constructFromPrePost(vector<int> &preorder, vector<int> &postorder) {
        for (int i = 0; i < postorder.size(); ++i) {
            postValToIdx[postorder[i]] = i;
        }

        return buildTree(preorder, 0, preorder.size() - 1, 
                         postorder, 0, postorder.size() - 1);
    }

private:
    unordered_map<int, int> postValToIdx;

    TreeNode *buildTree(vector<int> &preorder, int preHead, int preTail, 
                        vector<int> &postorder, int postHead, int postTail) {
        if (preHead > preTail) {
            return nullptr;
        }

        TreeNode *node = new TreeNode(preorder[preHead]);
        if (preHead == preTail) {
            return node;
        }

        int postLeftRoot = postValToIdx[preorder[preHead + 1]];
        int leftSize = postLeftRoot - postHead + 1;

        node->left = buildTree(preorder, preHead + 1, preHead + leftSize, 
                               postorder, postHead, postLeftRoot);
        node->right = buildTree(preorder, preHead + leftSize + 1, preTail,
                                postorder, postLeftRoot + 1, postTail - 1);
        return node;
    }
};

## Time Complexity
## Space Complexity
# Miscellaneous
[105. Construct Binary Tree from Preorder and Inorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)
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# Test Cases

[1,2,4,5,3,6,7]
[4,5,2,6,7,3,1]

[1]
[1]

[2,1]
[1,2]

-->