问题：
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.
解答：
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def longestUnivaluePath(self, root):
"""
:type root: TreeNode
:rtype: int
"""
"""
5(a)
/
5(b)
/ \
5(c) 5(d)
经过某一节点的最长的path，可能有三种情况
a-b-c
a-b-d
c-b-d
需要计算三种情况的最大值，
记录 a-b 的长度
记录 b-c 的长度
记录 b-d 的长度
"""
def path(root,val_dict):
if root.val not in val_dict:
val_dict.update({root.val:0})
if root.left==None and root.right==None:
return 0
# 左侧
left_val = 0
if root.left!=None:
if root.left.val == root.val:
left_val = path(root.left,val_dict)+1
else:
path(root.left,val_dict)
# 右侧
right_val = 0
if root.right!=None:
if root.right.val == root.val:
right_val = path(root.right,val_dict)+1
else:
path(root.right,val_dict)
# 更新dict
# 处理了多种情况，c-b-d形式
# 1) right_val==0,left_val!=0
# 2) right_val!=0,left_val==0
# 3) right_val!=0,left_val!=0
if root.val in val_dict and val_dict[root.val]< right_val+left_val:
val_dict[root.val] = right_val+left_val
# 父亲节点需要这个数字 a-b-d 或者 a-b-c 形式
return max(left_val,right_val)
val_dict = {}
if root!=None:
path(root,val_dict)
return max(val_dict.values())
return 0