LeetCode124 Binary Tree Maximum Path Sum

Given a binary tree, find the maximum path sum. （Hard）

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.

分析：

问题是从树种任意点到任意点的最大path sum，且至少有一个点在路径上。

先考虑一个简单的问题，从顶点到任意点的最短路径（可以不包含任何点，即为0）

这样可以用一个简单的递归实现：

int singleSum(TreeNode* root) {
        if (root == nullptr) {
            return 0;
        }
        return max(0,root -> val + max(singleSum(root -> left), singleSum(root -> right)));
}

再考虑pathSum与singleSum的关系；

pathSum采用分治的思想，可以求左子树的pathSum和右子树的pathSum，最后剩下的部分就是跨越根节点的pathSum

实际上也就是root->val加上左右子树的singlePath之和（如果小于0就不要了）。

程序：

 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     int singleSum(TreeNode* root) {
12         if (root == nullptr) {
13             return 0;
14         }
15         return max(0,root -> val + max(singleSum(root -> left), singleSum(root -> right)));
16     }
17 public:
18     int maxPathSum(TreeNode* root) {
19         if (root == nullptr) {
20             return -0x7FFFFFFF;
21         }
22         int leftSum = maxPathSum(root -> left);
23         int rightSum = maxPathSum(root -> right);
24         int midSum = root -> val + max(0, (singleSum(root -> left) + singleSum(root -> right)));
25         return max(midSum,max(leftSum, rightSum));
26     }
27 };

但是一组大样例居然超时了，考虑一下重复计算确实不少，给singleSum加一个hash表即可。

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     unordered_map<TreeNode* ,int> hash;
12     int singleSum(TreeNode* root) {
13         if (root == nullptr) {
14             return 0;
15         }
16         if (hash.find(root) != hash.end()) {
17             return hash[root];
18         }
19         int result = max(0,root -> val + max(singleSum(root -> left), singleSum(root -> right)));
20         hash[root] = result;
21         return result;
22     }
23 public:
24     int maxPathSum(TreeNode* root) {
25         if (root == nullptr) {
26             return -0x7FFFFFFF;
27         }
28         int leftSum = maxPathSum(root -> left);
29         int rightSum = maxPathSum(root -> right);
30         int midSum = root -> val + max(0, (singleSum(root -> left) + singleSum(root -> right)));
31         return max(midSum,max(leftSum, rightSum));
32     }
33 };