# Construct Binary Tree from Inorder and Postorder Traversal

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

> Given inorder and postorder traversal of a tree, construct the binary tree.

## Thoughts

和上题思路一致，只是这次root是先从postorder的末尾找。

## Code

```
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private TreeNode helper(int[] inorder, int instart, int inend, int[] postorder, int pstart, int pend) {
        if (instart > inend || pstart > pend) {
            return null;
        }
        int nodeVal = postorder[pend];
        int inpos = -1;
        // 0, 6 - 0
        for (int i = 0; i <= inend - instart; i++) {
            if (inorder[instart + i] == nodeVal) {
                inpos = i;
            }
        }
        TreeNode node = new TreeNode(inorder[instart + inpos]);
        node.left = helper(inorder, instart, instart + inpos - 1, postorder, pstart, pstart + inpos - 1);
        node.right = helper(inorder, instart + inpos + 1, inend, postorder, pstart + inpos, pend - 1);
        return node;
    }

    public TreeNode buildTree(int[] inorder, int[] postorder) {
        return helper(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);
    }
}
```

## Analysis

Errors:

1. inorder\[instart + i] == nodeVal忘了写instart

   做题耗时: 15min

时间复杂度O(n^2).

## Ver.2

和preorder-inorder同理, 只不过要倒过来处理. 从尾部遍历先建右树再建左树. 时间复杂度O(N).

```
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if (postorder.length == 0) {
            return null;
        }

        Stack<TreeNode> stack = new Stack<>();        
        TreeNode root = new TreeNode(postorder[postorder.length - 1]), cur = root;
        for (int i = postorder.length - 2, j = inorder.length - 1; i >= 0; i--) {
            if (cur.val != inorder[j]) {
                cur.right = new TreeNode(postorder[i]);
                stack.push(cur);
                cur = cur.right;
            } else {
                j--;
                while (!stack.isEmpty() && stack.peek().val == inorder[j]) {
                    cur = stack.pop();
                    j--;
                }
                cur = cur.left = new TreeNode(postorder[i]);
            }
        }
        return root;
    }
}
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://hao-fu-1.gitbook.io/oj/binary_tree_and_divide_conquer/divide-and-concquer/construct-binary-tree-from-inorder-and-postorder-traversal.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.