The path may startandend at any node in the tree. For example: Given the below binary tree, 1 / \ 2 3 Return 6. 2.解法分析: leetcode中给出的函数头为:int maxPathSum(TreeNode *root) 给定的数据结构为: Definitionforbinary tree *structTreeNode { *intval; * TreeNode *left; * TreeNo...
}intcurMax(TreeNode *root) {if(!root)return0;returnroot -> val + oneSide(root -> left) + oneSide(root ->right); }//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_...
:type root: TreeNode :rtype: int """ self.maxSum = float('-inf') 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 el...
取其4个值中得最大值作为子树的最大路径和。 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...
Binary Search Tree Iterator 74 -- 1:31 App Leetcode-0081. Search in Rotated Sorted Array II 84 -- 2:53 App Leetcode-0144. Binary Tree Preorder Traversal 63 -- 4:35 App Leetcode-0101. Symmetric Tree 7 -- 1:29 App Leetcode-0167. Two Sum II - Input Array Is Sorted 12...
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。
上面的一,二,三这几种情况都可以作为树的一个子树再计算,但第四种是不能作为一个子树再计算的,比如下面的 3,2,4组成的子树是不能作为1的子节点再进行计算的, privateint maxValue=Integer.MIN_VALUE;publicintmaxPathSum(TreeNoderoot){maxPathSumHelper(root);returnmaxValue;}publicintmaxPathSumHelper(Tree...
题目 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...
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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...}