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 must contain at least one node and does not need to go through the root.

For example:
Given the below binary tree,

       1
      / \
     2   3

Return 6. 

解析

• 这一题有点类似一维数组求最大子序列的和，一维最大子序列和从一维的角度判断，这里二叉树有左右子树考虑。help（root）方法的返回值的意思就是求得root节点到该root子孙的任意一节点的最大路径值（注意区别这里root和根节点root,这里root是递归下去的节点）。在这个函数里还比较了以root为路径节点

class Solution_124 {

    int sum = INT_MIN;

    int help(TreeNode* root)
    {
        if (!root)
        {
            return 0;
        }
        int left = max(0, help(root->left));
        int right = max(0, help(root->right));

        sum = max(sum, left + right + root->val); //包括当前节点在路径上的最大值

        return max(left, right) + root->val; //返回以当前节点为根的路径最大值
    }

public:
    int maxPathSum(TreeNode *root) {  //考虑有负节点情况

        if (!root)
        {
            return 0;
        }

        help(root);

        return sum;
    }
};

题目来源

• 124. Binary Tree Maximum Path Sum