[leetcode]Binary Tree Maximum Path Sum

二叉樹,找出任意一點到另一點的路徑,使得和最大.

開始sb看錯題了...

其實嘛...我們遞歸的來看...

如果只是一個節點,那么當然就是這個節點的值了.

如果這個作為root,那么最長路應該就是..

F(left) + F(right) + val...當然如果left,或者right<0就不用加了的= =

我盟從下往上找...

如果不這個不是root,那么就不能把left和right加起來了...因為只是一條路...

/**
 * 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 ans;
    int scanT(TreeNode* root){
        if(root == NULL) return 0;
        int left = scanT(root -> left);
        int right = scanT(root -> right);
        int val = root -> val;
        if(left > 0) val += left;
        if(right > 0) val += right;
        if(val > ans) ans = val;
        return max(root->val ,max(left +  root -> val , right + root -> val));
    }
    int maxPathSum(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(root == NULL) return 0;
        ans = root -> val;
        scanT(root);
        return ans;
    }
};