0298._Binary_Tree_Longest_Consecutive_Sequence.md

298. Binary Tree Longest Consecutive Sequence

难度: Medium

刷题内容

原题连接

• https://leetcode.com/problems/binary-tree-longest-consecutive-sequence/

内容描述

Given a binary tree, find the length of the longest consecutive sequence path.

The path refers to any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The longest consecutive path need to be from parent to child (cannot be the reverse).

Example 1:

Input:

   1
    \
     3
    / \
   2   4
        \
         5

Output: 3

Explanation: Longest consecutive sequence path is 3-4-5, so return 3.
Example 2:

Input:

   2
    \
     3
    / 
   2    
  / 
 1

Output: 2 

Explanation: Longest consecutive sequence path is 2-3, not 3-2-1, so return 2.

解题方案

思路 1 - 时间复杂度: O(N)- 空间复杂度: O(N)******

Top down

beats 94.30%

class Solution:
    def longestConsecutive(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        def helper(node, parent, length):
            if not node:
                return 
            length = length + 1 if parent and node.val == parent.val + 1 else 1
            self.max_len = max(self.max_len, length)
            helper(node.left, node, length)
            helper(node.right, node, length)
            
        self.max_len = 0
        helper(root, None, 0)
        return self.max_len

思路 2 - 时间复杂度: O(N)- 空间复杂度: O(N)******

Bottom Up

beats 54.37%

class Solution:
    def longestConsecutive(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        def helper(node):
            if not node:
                return 0
            L = helper(node.left) + 1
            R = helper(node.right) + 1
            if node.left and node.val + 1 != node.left.val:
                L = 1
            if node.right and node.val + 1 != node.right.val:
                R = 1
            length = max(L, R)
            self.max_len = max(self.max_len, length)
            return length
            
        self.max_len = 0
        helper(root)
        return self.max_len