3. R-sub + Node 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
}//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 ->...
int path_sum = max(singlePathSum(root->left), singlePathSum(root->right)); return max(0, (root->val + path_sum)); } }; 源码分析 注意singlePathSum中最后的返回值,如果其路径长度path_sum比0还小,那么取这一段路径反而会减少最终的路径和,故不应取这一段,我们使用0表示这一隐含意义。 题解...
self._maxPathSum(root) return self.maxSum def _maxPathSum(self, root): # DFS if root is None: return 0 left = self._maxPathSum(root.left) right = self._maxPathSum(root.right) left = left if left > 0 else 0 right = right if right > 0 else 0 self.maxSum = max(self.maxS...
class Solution { public: int maxPathSum(TreeNode* root) { int res = INT_MIN; maxDown(root, res); return res; } int maxDown(TreeNode* node, int& res) { if(!node) { return 0; } int left = max(0, maxDown(node->left, res)); // prune negative branch int right = max(0,...
[LeetCode][Java] Binary Tree Maximum Path Sum 题目: 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....
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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