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26. Binary Tree Maximum Path Sum
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
Return 6.
思想： 后序遍历。注意路径的连通： 结点不为空时要返回 max( max(leftV, rightV)+rootV, rootV);
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: int maxPathSum(TreeNode *root) { getMaxSum(root); return _maxPathSum; }protected: int getMaxSum(TreeNode *root) { if(root == NULL) return Min_Val; int left = getMaxSum(root->left); int right = getMaxSum(root->right); return findMax(left, right, root->val); } int findMax(int left, int right, int rootV) { int PathSum = max(max(left, right)+rootV, rootV); _maxPathSum = max(_maxPathSum, max(PathSum, rootV + left + right)); return PathSum; }private: enum{ Min_Val = -1000}; int _maxPathSum = Min_Val;};
26. Binary Tree Maximum Path Sum
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