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 };