【Binary Tree Maximum Path Sum】cpp

题目：

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.

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 maxPathSum(TreeNode* root){int max_sum = INT_MIN;Solution::dfs4MaxSum(root, max_sum);return max_sum;}static int dfs4MaxSum(TreeNode* root, int& max_sum){int ret=0;if ( !root ) return ret;int l = Solution::dfs4MaxSum(root->left, max_sum);int r = Solution::dfs4MaxSum(root->right, max_sum);if ( l>0 ) ret += l;if ( r>0 ) ret += r;ret += root->val;max_sum = std::max(ret, max_sum);return std::max(root->val, std::max(root->val+l, root->val+r));}
};

tips：

求array最大子序列和是一样的思路。只不过binary tree不是一路，而是分左右两路，自底向上算。

主要注意的地方如下：

1. 维护一个全局最优变量(max_sum)，再用ret存放包括前节点在内的局部最大值；每次比较max_sum和ret时，max_sum代表的是不包含当前节点在内的全局最优，ret代表的是一定包含当前节点在内的全局最优。

2. dfs4MaxSum返回时需要注意，left和right只能算一路，不能一起都返到上一层。因为题目中要求的path抻直了只能是一条线，不能带分叉的。

==============================================

第二次过这道题，大体思路有了，细节的错误也都犯了，改过来强化一次AC了。

/*** 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 maxPathSum(TreeNode* root){int ret = INT_MIN;if (root) Solution::maxPS(root, ret);return ret;}static int maxPS( TreeNode* root, int& maxSum){if ( !root ) return 0;int l = Solution::maxPS(root->left, maxSum);int r = Solution::maxPS(root->right, maxSum);maxSum = max(maxSum, root->val+(l>0?l:0)+(r>0?r:0));return max(root->val,root->val+max(r,l));}
};

转载于:https://www.cnblogs.com/xbf9xbf/p/4511277.html

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