题目
原题地址
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.
思路
用递归解决，从root节点开始遍历所有子节点，不走回头路保证时间复杂度为O(n)。最长路径可能经过root，但是也可能是在子节点中，所以要有一个全局变量储存最长路径。
考虑一个节点：
- 计算以它开始向左或向右的最长路径，作为递归返回值传给父节点
- 计算经过它的最长路径（可以同时向左和向右延伸），与全局最长路径比较
python代码
# 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 search(self, node):
if node == None:
return 0
num1 = self.search(node.left) # 从左子节点开始的最长路径
num2 = self.search(node.right) # 从右子节点开始的最长路径
if node.left and node.left.val == node.val:
num1 += 1 # 左子节点和节点值相同，从左子节点开始的最长路径也包含了本节点，所以+1
else:
num1 = 0 # 如果节点和左子节点值不同，从这个节点向左的路径长度就是0
if node.right and node.right.val == node.val:
num2 += 1
else:
num2 = 0
self.longest = max(self.longest, num1 + num2) # 经过此节点的最长路径与全局最长路径比较
return max(num1, num2) # 返回从此节点开始的最长路径
def longestUnivaluePath(self, root):
"""
:type root: TreeNode
:rtype: int
"""
self.longest = 0 # 全局最长路径
self.search(root)
return self.longest
