Leetcode解题-Construct Binary Tree from Inorder and Postorder Traversal

发表于   |   分类于   |  

描述

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

Note:
You may assume that duplicates do not exist in the tree.

分析

上一题的姐妹版。如法炮制即可。

PS: 从preorder和postorder是无法重建二叉树的。

代码

Python

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
class Solution(object):
    def buildTreeR(self, inorder, x1, y1, postorder, x2, y2):
        if x1 > y1:
            return None
        if x2 > y2:
            return None

        node = TreeNode(postorder[y2])
        root_idx = inorder.index(postorder[y2], x1, y1 + 1)
        left_size = root_idx - x1
        node.left = self.buildTreeR(inorder, x1, root_idx - 1,
                                    postorder, x2, x2 + left_size - 1)
        node.right = self.buildTreeR(inorder, root_idx + 1, y1,
                                     postorder, x2 + left_size, y2 - 1)
        return node

    def buildTree(self, inorder, postorder):
        """
        :type inorder: List[int]
        :type postorder: List[int]
        :rtype: TreeNode
        """
        return self.buildTreeR(inorder, 0, len(inorder) - 1,
                               postorder, 0, len(postorder) - 1)

热评文章