}//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
4. L-sub + Node + R-sub (in here the max value cannot be passed to the parent) publicstaticintmaxPathSum(TreeNode root){int[] max = {Integer.MIN_VALUE};// 因为Java里都是pass by value所以利用数组传!rec(root, max);returnmax[0]; }publicstaticintrec(TreeNode root,int[] max){if...
int maxPathSum(TreeNode* root) { int sum = INT_MIN; maxPathSum(root,sum); return sum; } int maxPathSum(TreeNode* root,int& sum){ if(!root) return 0; int sum1 = sum,sum2 = sum; int left = max(0,maxPathSum(root->left,sum1)); int right = max(0,maxPathSum(root->right...
1. When dfs method return, it returns the path with larger sum - only one path: could be root + left, and root + right 2. When checking current max sum path, left\right\root should be involved all, because such a path could be anything....
left); //右子节点的值 int right = maxPathSumHelper(root.right); //第4种情况 int cur = root.val + Math.max(0, left) + Math.max(0, right); //第1,2,3三种情况,返回当前值加上左右子节点的最大值即可,如果左右子节点都 //小于0,还不如不选 int res = root.val + Math.max(0, ...
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。
1191 K-Concatenation Maximum Sum K 次串联后最大子数组之和 Description: Given an integer array arr and an integer k, modify the array by repeating it k times. For example, if arr = [1, 2] and k = 3 then the modified array will be [1, 2, 1, 2, 1, 2]. ...
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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 needto go through the root. 举例: Gi...