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; * TreeNode *right; * TreeNode(intx) : val(x), left(NULL), right(NULL...
}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_...
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 root...
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...
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, AI检测代码解析 1 / \ 2 3 1. 2. 3. Return6. 题意: 给定一棵二叉树,找出最大得路径和。
LeetCode刷题日记 Day 8 Part 2 - Lowest Common Ancestor of a Binary Search Tree 83 -- 4:22 App LeetCode刷题日记 Day 17 Part 1 - Longest Common Prefix 2 -- 7:37 App LeetCode刷题日记 Day 91 Part 1 - Path with Maximum Probability 88 -- 5:13 App LeetCode刷题日记 Day 14 Part...
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。
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 root...
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。
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