### 687. Longest Univalue Path


题目:
<https://leetcode.com/problems/longest-univalue-path/>


难度:

Easy



思路

这道题也只能算个```easy```题目，根据传进来的```root```，我们只要从它的左右节点不停的递归下去，只要其```value```值与```root```一样，
该方向上的```length```就加```1```，最后我们将左右方向上的```length```相加, 递归取最大值
##### 很重要的一点就是，Note: The length of path between two nodes is represented by the number of edges between them.
- 因此是```self.res = max(self.res, left_arrow + right_arrow)```, ```return max(left_arrow, right_arrow)```
- 而不是```self.res = max(self.res, left_arrow + right_arrow + 1)```, ```return max(left_arrow + 1, right_arrow + 1)```





```python
# 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.res = 0
        def dir_length(node):
            if not node:
                return 0
            left_len = dir_length(node.left)     # 左节点的length
            right_len = dir_length(node.right)   # 右节点的length
            left_dir, right_dir = 0, 0
            if node.left and node.left.val == node.val:
                left_dir = left_len + 1          # 当前节点的左节点方向的length
            if node.right and node.right.val == node.val:
                right_dir = right_len + 1        # 当前节点的右边节点方向的length
            self.res = max(self.res, left_dir + right_dir)
            return max(left_dir, right_dir)
        dir_length(root)
        return self.res
```



