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...
* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public:intmaxPathSum(TreeNode*root) { rec(root);returnmax; }intrec(TreeNode *root) {if(root ==NULL)return0;intl = rec(root->left);intr = rec(root->right);if(root->val>max) max = root->val;...
class Solution: ans = float('-inf') def maxPathSum(self, root: TreeNode) -> int: def helper(node): if not node: return 0 l = helper(node.left) r = helper(node.right) self.ans = max(self.ans, max(l,0) + max(r, 0) + node.val) return max(l, r...
# self.left=None # self.right=NoneclassSolution:res=-sys.maxsize defmaxPathSum(self,root:TreeNode)->int:self.dfs(root)returnself.res defdfs(self,root:TreeNode)->int:ifroot==None:return0l_max_sum=self.dfs(root.left)r_max_sum=self.dfs(root.right)sum=root.val sum=max(sum,sum+l_m...
二叉树中的最大路径和 ---> [不考虑负数问题]-->叉树中的最长路径-->从root节点到叶子节点肯定不是,具体哪个还不知道,肯定有一个】---> root+max(left)+max(right)在遍历方式: 一个路径:选择其中一个节点 为中心节点,然后 左右2边累计结果 路径的问题:变成高度的问题 。遍历方式保存还是不变。 他山之...
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 ...
classSolution {public:intmaxSum =INT_MIN;intgetMaxSum(TreeNode*root){if(root == NULL)return0;intleftSum = getMaxSum(root->left), rightSum = getMaxSum(root->right);intcurSum = max(root->val, max(root->val+leftSum, root->val+rightSum)); ...
(node.val+max(self.maxPath(node.left),self.maxPath(node.right)),node.val)returnnode.maxPdefmaxPathSum(self,root:TreeNode)->int:ifnotroot:return-sys.maxsize-1#调用一次步骤1,求出并存储所有节点的最大路径和self.maxPath(root)#第二次递归,求每个几点为根节点的最大路径和(题目所定义的),返回...
def maxPathSum(self, root): ''' :type root: TreeNode :rtype: int ''' self.maxes = -float('inf')def helper(root): myValue = root.valif not root.left and not root.right: self.maxes = myValue if myValue > self.maxes else self.maxes ...
_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.maxSum, ...