106.Construct Binary Tree from Inorder and Postorder Traversal
===

###### tags: `Medium`,`Tree`,`Array`,`Hash Table`,`Divide and Conquer`

[106. Construct Binary Tree from Inorder and Postorder Traversal](https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/)

### 題目描述

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*.

### 範例

**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.

### 解答
#### Python

class Solution:
    def buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
        idx_map = {val:idx for idx, val in enumerate(inorder)}

        def recur(low, high):
            if low > high:
                return

            idx = idx_map[postorder.pop()]
            
            root = TreeNode(inorder[idx])
            root.right = recur(idx + 1, high)
            root.left = recur(low, idx - 1)

            return root

        return recur(0, len(inorder) - 1)

> [name=Ron Chen][time= Thu, Mar 16, 2023]

#### Javascript

function buildTree(inorder, postorder) {
  const root = postorder[postorder.length - 1];
  const rootIndex = inorder.indexOf(root);
  const leftInorder = inorder.slice(0, rootIndex);
  const rightInorder = inorder.slice(rootIndex + 1);

  const leftPostorder = postorder.slice(0, leftInorder.length);
  const rightPostorder = postorder.slice(leftInorder.length, postorder.length - 1);

  const tree = new TreeNode(root);
  if (leftInorder.length > 0) {
    tree.left = buildTree(leftInorder, leftPostorder);
  }

  if (rightInorder.length > 0) {
    tree.right = buildTree(rightInorder, rightPostorder);
  }

  return tree;
}

> [name=Marsgoat][time=Mar 17, 2023]

### Reference

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