leetcode || 124、Binary Tree Maximum Path Sum

problem：

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.

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题意：在一棵二叉树中寻找一条路径，使其和最大

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);
    }
};