}//对每个节点遍历求左右两个节点的做大加上本身,然后取最大的值就是maximum path sum了intmaxPathSum(TreeNode *root) {if(!root)return0;inttmpl = INT_MIN, tmpr =INT_MIN;intcur =curMax(root);if(root ->left) tmpl= maxPathSum(root ->left);if(root ->right) tmpr= maxPathSum(root ->...
接着我们发现上述计算max 和 求出MAX的过程完全可以放到func(node) 里去。 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), r...
Leetcode Binary Tree Maximum Path Sum Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. For example: Given the below binary tree, 1 / \ 2 3 1. 2. 3. Return6. 对于每个结点,我们需要比较三个值。 左结点为结尾的路径和+当前结点值 ...
* Definition for a binary tree node. * public class TreeNode{* int val; * TreeNode left; * TreeNode right; * TreeNode(int x){val = x;}*}*/publicclassSolution{publicintmaxPathSum(TreeNoderoot){intmax[]=newint[1];max[0]=Integer.MIN_VALUE;calculateSum(root,max);returnmax[0];}pu...
LeetCode刷题日记 Day 8 Part 2 - Lowest Common Ancestor of a Binary Search Tree 83 -- 4:22 App LeetCode刷题日记 Day 17 Part 1 - Longest Common Prefix 2 -- 7:37 App LeetCode刷题日记 Day 91 Part 1 - Path with Maximum Probability 88 -- 5:13 App LeetCode刷题日记 Day 14 Part...
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。 ** 总结: pre-order -> top-down post-order -> bottom-up in-order -> ? level-order -> bfs ...
Given a non-empty 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...
题目描述: {代码...} connections. The path must contain at least one node and does not needto go through the root. 举例: {代码...}
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Binary Tree Maximum Path Sum@LeetCode Binary Tree Maximum Path Sum 动态规划+深度优先搜索。把大问题(求整棵树的路径最大值)拆分成小问题(每颗子树的路径最大值),递推公式为:当前树的路径最大值=max(左子树的路径最大值, 右子树的路径最大值)+当前根节点的值。以此来推出最后全树的最大路径值。