Given a 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 does not need to go through the root.
For example:
Given the below binary tree,
1
/
2 3
Return 6.
思路：所求的maximum path sum有可能会包括一个根节点和两条分别在左右子树中的路径，这里构建一个新的树，每个节点存储的值为所给二叉树对应位置节点向下的最大单条路径(不会在根节点分叉)。然后遍历整个树，找到maximum path sum。
代码中copyTree函数为构建树。遍历树并找到maximum path sum在help函数中实现。
1 class Solution {
2 public:
3 TreeNode* copyTree(TreeNode* root)
4 {
5 if (!root) return NULL;
6 TreeNode* cur = new TreeNode(root->val);
7 cur->left = copyTree(root->left);
8 cur->right = copyTree(root->right);
9 int largest = 0;
10 if (cur->left) largest = max(largest, cur->left->val);
11 if (cur->right) largest = max(largest, cur->right->val);
12 cur->val += largest;
13 return cur;
14 }
15 void help(TreeNode* root, TreeNode* cproot, int& res)
16 {
17 if (!root) return;
18 int localSum = root->val;
19 if (cproot->left && cproot->left->val > 0)
20 localSum += cproot->left->val;
21 if (cproot->right && cproot->right->val > 0)
22 localSum += cproot->right->val;
23 if (localSum > res) res = localSum;
24 help(root->left, cproot->left, res);
25 help(root->right, cproot->right, res);
26 }
27 int maxPathSum(TreeNode* root) {
28 TreeNode* cproot = copyTree(root);
29 int res = INT_MIN;
30 help(root, cproot, res);
31 return res;
32 }
33 };