[LeetCode] 124. Binary Tree Maximum Path Sum

Binary Tree Maximum Path Sum

Given a non-empty 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.

解析

构建要给helper辅助函数，在递归函数中，如果当前结点不存在，那么直接返回0。否则就分别对其左右子节点调用递归函数，由于路径和有可能为负数，而我们当然不希望加上负的路径和，所以我们和0相比，取较大的那个，就是要么不加，加就要加正数。然后我们来更新全局最大值结果res，就是以左子结点为终点的最大path之和加上以右子结点为终点的最大path之和，还要加上当前结点值，这样就组成了一个条完整的路径。而我们返回值是取left和right中的较大值加上当前结点值，因为我们返回值的定义是以当前结点为终点的path之和，所以只能取left和right中较大的那个值，而不是两个都要。

代码

class Solution {
public:
    int maxPathSum(TreeNode* root) {
        int res = INT_MIN;
        helper(root, res);
        return res;
    }
    int helper(TreeNode* root, int& res){
        if(!root) return 0;
        int left = max(helper(root->left, res), 0);
        int right = max(helper(root->right, res), 0);
        res = max(res, left + right + root->val);
        return max(left, right) + root->val;
    }
};

参考

http://www.cnblogs.com/grandyang/p/4280120.html