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.
Solution: Recursion...
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 int maxPathSum(TreeNode *root) { 13 int res = INT_MIN; 14 maxPathSumRe(root, res); 15 return res; 16 } 17 18 int maxPathSumRe(TreeNode *root, int &res) 19 { 20 if(!root) return 0; 21 int l = maxPathSumRe(root->left, res); 22 int r = maxPathSumRe(root->right, res); 23 int sum = max(root->val, max(l,r) + root->val); // max (root as node) 24 res = max(sum, res); 25 res = max(root->val + l + r, res); 26 return sum; 27 } 28 };