标签:type ini index not none tor def one build
根据一棵树的中序遍历与后序遍历构造二叉树。
中序遍历 inorder = [9,3,15,20,7]
后序遍历 postorder = [9,15,7,20,3]
3
/ 9 20
/ 15 7
Python 实现
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def buildTree(self, inorder, postorder):
"""
:type inorder: List[int]
:type postorder: List[int]
:rtype: TreeNode
"""
if not postorder or not inorder:
return None;
node = TreeNode(postorder[-1])
mid = inorder.index(postorder[-1])
node.left = self.buildTree(inorder[:mid],postorder[:mid])
node.right = self.buildTree(inorder[mid+1:],postorder[mid:-1])
return node
标签:type ini index not none tor def one build
原文地址:https://www.cnblogs.com/xmxj0707/p/9694731.html