Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root.

Note: The length of path between two nodes is represented by the number of edges between them.

Example 1:

Input:

              5
             / \
            4   5
           / \   \
          1   1   5
 

Output:

2

Example 2:

Input:

              1
             / \
            4   5
           / \   \
          4   4   5

Output:

2

Note: The given binary tree has not more than 10000 nodes. The height of the tree is not more than 1000.

Runtime: 72 ms, faster than 28.47% of C++ online submissions for Longest Univalue Path.

對於這種不經過root的求和題往往都需要一個臨時變數，然後考慮一下根節點和子節點的關係，用一個引用得到最優解。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
  int longestUnivaluePath(TreeNode* root) {
    
 
int ret = 0, tmpret = 0;
    helper(root,tmpret, ret);
    return ret == 0 ? 0 : ret - 1;
  }
  int helper(TreeNode* root, int& tmpret, int& ret){
    if(!root) return 0;
    int rl = helper(root->left, tmpret, ret);
    int rr = helper(root->right, tmpret, ret);
    if(!root->left && !root->right) {
      tmpret = 1;
      //ret = 1;
      return 1;
    } else if(!root->left && root->right){
      if(root->val == root->right->val){
        tmpret = max(tmpret, rr + 1);
        ret = max(ret, tmpret);
        return 1+rr;
      } else return 1;
    } else if(root->left && !root->right){
      if(root->val == root->left->val){
        tmpret = max(tmpret, 1+rl);
        ret = max(ret, tmpret);
        return 1+rl;
      } else return 1;
    } else {
      if(root->val == root->left->val && root->val == root->right->val){
        tmpret = max(tmpret, 1+rr + rl);  
        ret = max(ret, tmpret);
        return 1+max(rr,rl);
      } else if (root->val == root->left->val){
        tmpret = max(tmpret, 1+rl);
        ret = max(ret, tmpret);
        return 1+rl;
      } else if (root->val == root->right->val){
        tmpret = max(tmpret, 1+rr);
        ret = max(ret, tmpret);
        return 1+rr;
      } else return 1;
    }
  }
};