[LeetCode] Longest Univalue Path

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：

1. 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
2. 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