题目

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

解题思路

跟LeetCode: 105. Construct Binary Tree from Preorder and Inorder Traversal一模一样的解法，只是把前序数组的每次输出第一个换成后序数组输出最后一个。此外，105是先左节点再右节点，106反过来。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
#         while postorder:
#             print(postorder.pop())if inorder:root_val = postorder.pop()# print(inorder.index(root_val))ind = inorder.index(root_val)root = TreeNode(root_val)root.right = self.buildTree(inorder[ind+1:], postorder)root.left = self.buildTree(inorder[0:ind], postorder)return root

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