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(root ==null){return0; }intleftSubtreeMaxSum=rec(root.left, max);// 左子树的最大和...
}intmaxPathSumCore(TreeNode *node) {if(NULL == node)return0;inta = maxPathSumCore(node ->left);intb = maxPathSumCore(node ->right);if((a+b+node->val) > MAX) MAX = (a+b+node->val);if((a+node->val) > MAX) MAX = (a+node->val);if((b+node->val) > MAX) MAX = ...
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。 实现...
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.maxSum, root.val + left +...
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。
参考: https://shenjie1993.gitbooks.io/leetcode-python/124%20Binary%20Tree%20Maximum%20Path%20Sum.html 我们现在要求最大路径和,那么就要分别得到左右两条路径的最大和。而左路径的最大和为左节点的值加上它左右路径中较大的路径和,右路径最大和为右节点的值加上它左右路径中较大的路径和。 注意:如果...
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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