# 0106. Construct Binary Tree from Inorder and Postorder Traversal

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

## Description

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:**

![](https://assets.leetcode.com/uploads/2021/02/19/tree.jpg)

```
**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.

## ac

```java
/**
 * 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) {
        // edge cases
        if (postorder.length == 0 || inorder.length == 0) return null;

        return helper(inorder, 0, inorder.length-1, postorder, 0, postorder.length-1);
    }
    private TreeNode helper(int[] inorder, int istart, int iend, int[] postorder, int pstart, int pend) {
        // exit
        if (pstart > pend || istart > iend) return null;
        if (pstart == pend || istart == iend) return new TreeNode(postorder[pstart]);

        // find root index
        int rootIdx = -1;
        for (int i = istart; i <= iend; i++) {
            if (postorder[pend] == inorder[i]){
                rootIdx = i;
                break;
            }
        }
        int len = rootIdx - istart;

        // divide and conquer
        TreeNode root = new TreeNode(inorder[rootIdx]);
        root.left = helper(inorder, istart, rootIdx-1, postorder, pstart, pstart+len-1);
        root.right = helper(inorder, rootIdx+1, iend, postorder, pstart+len, pend-1);

        return root;
    }
}
```


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