LeetCode（124） Binary Tree Maximum Path Sum

题目

Given a binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path does not need to go through the root.

For example:
Given the below binary tree,

       1
      / \
     2   3

 

Return 6.

分析

给定一颗二叉树，求其最大路径和。对于二叉树，算法大多可以选择递归解决，此题也不例外。

思路参考

如果只是一个节点,那么当然就是这个节点的值了.

如果这个作为root,那么最长路应该就是..

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

从下往上递归遍历...

如果不这个不是root,那么就不能把left和right加起来了...因为只是一条路...

AC代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    int maxVal = INT_MIN;
    int maxPathSum(TreeNode* root) {
        if (root == NULL)
            return 0;

        maxSum(root);
        return maxVal;
    }

    /*递归函数*/
    int maxSum(TreeNode *root)
    {
        if (root == NULL)
            return 0;

        /*求以root为根的当前子树的最大路径和*/
        int curVal = root->val;
        int lmaxSum = maxSum(root->left), rmaxSum = maxSum(root->right);
        if (lmaxSum > 0)
            curVal += lmaxSum;
        if (rmaxSum > 0)
            curVal += rmaxSum;

        if (curVal > maxVal)
            maxVal = curVal;

        /*返回以当前root为根的子树的最大路径和*/
        return max(root->val, max(root->val + lmaxSum, root->val + rmaxSum));
    }
};