題目：

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.

解釋： 題目很好理解，注意path的長度是邊的個數，path可以不經過root。 不經過root怎麼寫？？？ 和543. Diameter of Binary Tree（python+cpp）

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
 


class Solution(object):
    def longestUnivaluePath(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        self.max_path=0
        def TreeDepth(root):
            if root==None:
                return 0
            left_depth=0
            right_depth=0
            if root.left:
                left_depth=TreeDepth(root.left) if root.left.val==root.val else 0
            if root.right:
                right_depth=TreeDepth(root.right) if root.right.val==root.val else 0
            self.max_path=max(self.max_path,left_depth+right_depth)
            return max(left_depth,right_depth)+1
        TreeDepth(root)
        return self.max_path

正確解法： python程式碼：

 # Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def longestUnivaluePath(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        self.max_path=0
        def TreeDepth(root):
            left_depth=0
            right_depth=0
            if root.left:
                left_depth=TreeDepth(root.left)
                if root.left.val!=root.val:
                    left_depth=0
            if root.right:
                right_depth=TreeDepth(root.right)
                if root.right.val!=root.val:
                    right_depth=0
            self.max_path=max(self.max_path,left_depth+right_depth)
            return max(left_depth,right_depth)+1
        if root:
            TreeDepth(root)
        return self.max_path

c++程式碼：

/**
 * 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 max_path=0;
    int longestUnivaluePath(TreeNode* root) {
        if (root)
            TreeDepth(root);
        return max_path;
    }
    int TreeDepth(TreeNode *root)
    { 
        int left_depth=0;
        int right_depth=0;
        if(root->left)
        {
            left_depth=TreeDepth(root->left);
            if(root->left->val!=root->val)
                left_depth=0;
        }
        if(root->right)
        {
            right_depth=TreeDepth(root->right);
            if(root->right->val!=root->val)
                right_depth=0;
        }
        max_path=max(max_path,left_depth+right_depth);
        return max(left_depth,right_depth)+1;
    }
  
};

總結：