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Binary Tree Maximum Path Sum

124. Binary Tree Maximum Path Sum

题目链接:https://leetcode.com/problems/binary-tree-maximum-path-sum/#/description

题目大意:给定一棵二叉树,查找该二叉树的最大路径和。路径的定义是,从某个结点出发,沿着父子连接达到任意结点时经过的结点序列。路径至少包含一个结点,但不要求一定要经过根结点。

 

思路:对于根节点为root的树,以根节点为端点的最大路径和maxEndRoot(root)=max(root->val, maxEndRoot(root->left) + root->val, maxEndRoot(root->right) + root->val),不要求经过根结点的最大路径和maxPath(root) = max(maxPath(root->left), maxPath(root->right), maxEndRoot(root), maxEndRoot(root->left) + root->val + maxEndRoot(root->right))。

算法复杂度分析:时间复杂度O(n),空间复杂度O(log(n))。

代码:

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int maxPathSum(TreeNode* root) {
13         int maxSum = root->val;
14         recurMaxSum(root, maxSum);
15         return maxSum;
16     }
17 private:
18     int recurMaxSum(TreeNode *root, int &maxSum) {
19         if (!root)
20             return 0;
21         int sum = root->val;
22         int leftSum = recurMaxSum(root->left, maxSum);
23         sum += leftSum > 0 ? leftSum : 0;
24         int rightSum = recurMaxSum(root->right, maxSum);
25         sum += rightSum > 0 ? rightSum : 0;
26         maxSum = max(maxSum, sum);
27         return max(root->val, max(leftSum + root->val, rightSum + root->val));
28     }
29 };

评测系统上运行结果:

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Binary Tree Maximum Path Sum