leetcode || 124、Binary Tree Maximum Path Sum

problem：

       1
      /      2   3

thinking：

（1）二叉树寻找一条路径比较难做，没有parent指针更难

（2）采用DFS遍历，从根结点开始

code：

/**
 * 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 calLen(TreeNode *root, int &len)
    {
        if (root == NULL)
        {
            len = 0;
            return 0;
        }
        
        if (root->left == NULL && root->right == NULL)
        {
            len = root->val;
            return root->val;
        }
        
        int leftPath, rightPath;
        int leftLen;
        if (root->left)
            leftLen = calLen(root->left, leftPath);
        else
        {
            leftLen = INT_MIN;
            leftPath = 0;
        }
        
        int rightLen;
        if (root->right)
            rightLen = calLen(root->right, rightPath);
        else
        {
            rightLen = INT_MIN;
            rightPath = 0;
        }
        
        len = max(max(leftPath, rightPath) + root->val, root->val);
        int maxLen = max(root->val, max(leftPath + rightPath + root->val, 
            max(leftPath + root->val, rightPath + root->val)));
        
        return max(max(leftLen, rightLen), maxLen);
    }
    
    int maxPathSum(TreeNode *root) {
        int len;
        return calLen(root, len);
    }
};

leetcode || 124、Binary Tree Maximum Path Sum

原文：http://blog.csdn.net/hustyangju/article/details/45307843