按照这个思路的代码,maxPathSumCore 就是上面 func(node)的实现: /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public:intmaxPathSum(TreeNode *ro...
classSolution {public:intmaxPathSum(TreeNode *root) {//Start typing your C/C++ solution below//DO NOT write int main() functioninttmp=-99999999;if(!root)return0; maxPathSum2(root,tmp);returntmp; }intmaxPathSum2(TreeNode *root ,int&path){intlefttmp=-99999999;intrighttmp=-99999999;if(...
public int maxPathSum(TreeNode root) { if (root == null) { return Integer.MIN_VALUE; } //左子树的最大值 int left = maxPathSum(root.left); //右子树的最大值 int right = maxPathSum(root.right); //再考虑包含根节点的最大值 int all = ...; return Math.max(Math.max(left, right...
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 example: Given the below binary tree, ...
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 must contain at least one node and does not need to go through the root. ...
# Definition for a binary tree node.# class TreeNode(object):# def __init__(self, x):# self.val = x# self.left = None# self.right = NoneclassSolution(object):defmaxPathSum(self,root):""":typeroot:TreeNode:rtype:int"""self.maxSum=0self.findMax(root)returnself.maxSumdeffindMax...
http://bangbingsyb.blogspot.com/2014/11/leetcode-binary-tree-maximum-path-sum.html http://www.programcreek.com/2013/02/leetcode-binary-tree-maximum-path-sum-java/ 另外,这道题目的BST条件,似乎没什么用。因为如果全是负数,BST也没帮助了。
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The problem (MSPITH) aims to maximize the length of the shortest path from the root of a tree to all its leaves by upgrading edge weights such that the upgrade cost under sum-Hamming distance is upper-bounded by a given value. We show that the problem (MSPITH) under weighted sum-...
int maxPath = Math.max(left.maxPath, right.maxPath); 比如我们在计算rootToAny的时候,如下code: int singlePath = Math.max(left.singlePath, right.singlePath) + root.val; 可以帮助我们计算出。 所以我想请问下,为什么单独的Math.max();可以帮助我们计算出anyToAny? 谢谢老师答疑!添加...