Binary Tree Maximum Path Sum - LeetCode

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。

 class Solution {
 public:
     TreeNode* copyTree(TreeNode* root)
     {
         if (!root) return NULL;
         TreeNode* cur = new TreeNode(root->val);
         cur->left = copyTree(root->left);
         cur->right = copyTree(root->right);
         int largest = ;
         if (cur->left) largest = max(largest, cur->left->val);
         if (cur->right) largest = max(largest, cur->right->val);
         cur->val += largest;
         return cur;
     }
     void help(TreeNode* root, TreeNode* cproot, int& res)
     {
         if (!root) return;
         int localSum = root->val;
         if (cproot->left && cproot->left->val > )
             localSum += cproot->left->val;
         if (cproot->right && cproot->right->val > )
             localSum += cproot->right->val;
         if (localSum > res) res = localSum;
         help(root->left, cproot->left, res);
         help(root->right, cproot->right, res);
     }
     int maxPathSum(TreeNode* root) {
         TreeNode* cproot = copyTree(root);
         int res = INT_MIN;
         help(root, cproot, res);
         return res;
     }
 };