描述:递归调用,getMax返回 [节点值,经过节点左子节点的最大值,经过节点右节点的最大值],每次递归同时查看是否存在不经过节点的值大于max. 代码:待优化 def getLargeNode(self, a, b): if a and b: return max(a, b) elif a and not b: return a elif not a and b: return b else: tmp = None def getMax(self, node): if node is None…

题目: Given a binary tree, find the maximum path sum. For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path does not need to go through the root. For ex…

语言:Python 描述:使用递归实现 # Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: # @param p, a tree node # @param q, a tree node # @return a boolean def isSameTre…

语言:Python 描述:使用递归实现 class Solution: # @return an integer def numTrees(self, n): : elif n == : else: part_1 = self.numTrees(n-) * part_2 = ,n-): part_left = self.numTrees(i) part_right = self.numTrees(n - - i) part_2 += part_left * part_right return p…

题目: Invert a binary tree. 4 / \ 2 7 / \ / \ 1 3 6 9 to 4 / \ 7 2 / \ / \ 9 6 3 1 思路分析: 题意是将二叉树全部左右子数对调,如上图所看到的. 详细做法是,先递归处理左右子树,然后将当前的左右子树对调. 代码实现: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeN…

A balanced binary tree is something that is used very commonly in analysis of computer science algorithms. In this lesson we cover how to determine the maximum number of items it can accommodate. We follow this with a discussion on the maximum height…

Given a sorted (increasing order) array, write an algorithm to create a binary tree with minimal height. 1.Divide the array equally into left part and right part, the mid value will be the root. 2.Recall the function to the left and right part of the…

Problem A tree is a connected graph with no cycles. A rooted tree is a tree in which one special vertex is called the root. If there is an edge between X and Y in a rooted tree, we say that Y is a child of X if X is closer to the root than Y (in othe…

June 8, 2015 我最喜欢的一道算法题目, 二行代码. 编程序需要很强的逻辑思维, 多问几个为什么, 可不可以简化．想一想, 二行代码, 五分钟就可以搞定; 2015年网上大家热议的 Homebrew 的作者 Max Howell 面试 Google 挂掉的一题, 二叉树反转, 七行代码, 相比二行代码, 情有可原! Problem: return the count of binary tree with only one child 想一想, 你要写几行, 六七行, 或小于十行? S…

Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] 思路:用两个stack<TreeNode*> in , s; in : 记录当前的路径 p  , 和vector<…