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https://leetcode.com/problems/binary-tree-maximum-path-sum/
Given a 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 does not need to go through the root.
For example:
Given the below binary tree,
1 / 2 3
Return 6
.
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int res = INT_MIN; int dfs(TreeNode* root) { if(!root) return 0; int l = dfs(root->left); int r = dfs(root->right); int rhs = root->val; if(l > 0) rhs += l; if(r > 0) rhs += r; res = max(res, rhs); return max(l, r) <= 0? root->val: max(l, r) + root->val; } int maxPathSum(TreeNode* root) { if(!root) return 0; dfs(root); return res; } };
leetcode@ [124] Binary Tree Maximum Path Sum (DFS)
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原文地址:http://www.cnblogs.com/fu11211129/p/5098419.html