https://leetcode.com/problems/binary-tree-longest-consecutive-sequence/
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).
For example,
1
\
3
/ \
2 4
\
5
Longest consecutive sequence path is 3-4-5, so return 3.
2
\
3
/
2
/
1
Longest consecutive sequence path is 2-3,not3-2-1, so return 2.
Code
# 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 longestConsecutive(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if root is None:
return 0
self.res = 0
self.helper(root, root.left, 1)
self.helper(root, root.right, 1)
return self.res
def helper(self, prev, cur, length):
if cur is None:
self.res = max(self.res, length)
return
if cur.val == prev.val + 1:
self.helper(cur, cur.left, length+1)
self.helper(cur, cur.right, length+1)
else:
self.res = max(self.res, length)
self.helper(cur, cur.left, 1)
self.helper(cur, cur.right, 1)
Idea
There are three parameters of
helper function, prev (the previous node), cur (the current node) and length (the accumulated length before cur). If cur.val == prev.val + 1, then we can continue exploring the current node’s children with length increased by 1 (line 28, 29). Otherwise, refresh length. (line 32, 33)The time complexity in the worst case will always be O(N) where N is the number of tree nodes.
Of course you can implement one that does not require a global variable: https://discuss.leetcode.com/topic/29205/simple-recursive-dfs-without-global-variable