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binary-tree-longest-consecutive-sequence.py
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"""
# https://baihuqian.github.io/2018-08-01-binary-tree-longest-consecutive-sequence/
Leetcode 298: Binary Tree Longest Consecutive Sequence
Question
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.
"""
# V0
# IDEA : DFS
class Solution(object):
def longestConsecutive(self, root):
if not root:
return 0
self.result = 0
self.helper(root, 1)
return self.result
def helper(self, root, curLen):
self.result = curLen if curLen > self.result else self.result
if root.left:
if root.left.val == root.val + 1:
self.helper(root.left, curLen + 1)
else:
self.helper(root.left, 1)
if root.right:
if root.right.val == root.val + 1:
self.helper(root.right, curLen + 1)
else:
self.helper(root.right, 1)
# V0'
# IDEA : BFS
class Solution(object):
def longestConsecutive(self, root):
if root is None:
return 0
stack = list()
stack.append((root, 1))
maxLen = 1
while len(stack) > 0:
node, pathLen = stack.pop()
if node.left is not None:
if node.val + 1 == node.left.val:
stack.append((node.left, pathLen + 1))
maxLen = max(maxLen, pathLen + 1)
else:
stack.append((node.left, 1))
if node.right is not None:
if node.val + 1 == node.right.val:
stack.append((node.right, pathLen + 1))
maxLen = max(maxLen, pathLen + 1)
else:
stack.append((node.right, 1))
return maxLen
# V0''
# IDEA : BFS
# TODO : validate this approach
# BFS
# class Solution(object):
# def longestConsecutive(self, root):
# # edge case
# if not root:
# return 0
# cache = []
# q = [[cache, root]]
# res = []
# while q:
# for i in range(len(q)):
# cache, tmp = q.pop(0)
# if cache:
# if tmp.val = cache[-1] + 1:
# cache.append(tmp.val)
# res = max(res, len(cache))
# else:
# res = max(res, len(cache))
# cache = []
# if tmp.left:
# q.append([cache, tmp.left])
#
# if tmp.right:
# q.append([cache, tmp.right])
# return res
# V1
# https://www.jianshu.com/p/ebabdeed9bca
# IDEA : DFS
class Solution(object):
def longestConsecutive(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root:
return 0
self.result = 0
self.helper(root, 1)
return self.result
def helper(self, root, curLen):
self.result = curLen if curLen > self.result else self.result
if root.left:
if root.left.val == root.val + 1:
self.helper(root.left, curLen + 1)
else:
self.helper(root.left, 1)
if root.right:
if root.right.val == root.val + 1:
self.helper(root.right, curLen + 1)
else:
self.helper(root.right, 1)
### Test case : dev
# V1'
# IDEA : BFS
# https://baihuqian.github.io/2018-08-01-binary-tree-longest-consecutive-sequence/
class Solution(object):
def longestConsecutive(self, root):
if root is None:
return 0
stack = list()
stack.append((root, 1))
maxLen = 1
while len(stack) > 0:
node, pathLen = stack.pop()
if node.left is not None:
if node.val + 1 == node.left.val:
stack.append((node.left, pathLen + 1))
maxLen = max(maxLen, pathLen + 1)
else:
stack.append((node.left, 1))
if node.right is not None:
if node.val + 1 == node.right.val:
stack.append((node.right, pathLen + 1))
maxLen = max(maxLen, pathLen + 1)
else:
stack.append((node.right, 1))
return maxLen
# V1'
# https://eugenejw.github.io/2017/08/leetcode-298
# IDEA : DFS
class Solution(object):
def __init__(self):
self.ret = 0
def longestConsecutive(self, root):
"""
:type root: TreeNode
:rtype: int
:algorithm: Pre-order DFS, O(n)
:runtimr: 188ms
"""
if not root:
return 0
self.dfs(root, 1)
return self.ret
def dfs(self, root, carry):
"""
:type root: TreeNode
:rtype: void
"""
self.ret = max(self.ret, carry)
if root.left:
if root.left.val - 1 == root.val:
self.dfs(root.left, carry+1)
else:
self.dfs(root.left, 1)
if root.right:
if root.right.val - 1 == root.val:
self.dfs(root.right, carry+1)
else:
self.dfs(root.right, 1)
# V1
# https://blog.csdn.net/qq508618087/article/details/50883425
# JAVA
# /**
# * Definition for a binary tree node.
# * struct TreeNode {
# * int val;
# * TreeNode *left;
# * TreeNode *right;
# * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
# * };
# */
# class Solution {
# public:
# void DFS(TreeNode* root, int pre, int len, int& Max)
# {
# if(!root) return;
# len = (root->val==pre+1)?len+1:1;
# Max = max(Max, len);
# DFS(root->left, root->val, len, Max);
# DFS(root->right, root->val, len, Max);
# }
#
# int longestConsecutive(TreeNode* root) {
# if(!root) return 0;
# int Max = 1;
# DFS(root, root->val, 0, Max);
# return Max;
# }
# };
# V2
# Time: O(n)
# Space: O(h)
class Solution(object):
def longestConsecutive(self, root):
"""
:type root: TreeNode
:rtype: int
"""
self.max_len = 0
def longestConsecutiveHelper(root):
if not root:
return 0
left_len = longestConsecutiveHelper(root.left)
right_len = longestConsecutiveHelper(root.right)
cur_len = 1
if root.left and root.left.val == root.val + 1:
cur_len = max(cur_len, left_len + 1)
if root.right and root.right.val == root.val + 1:
cur_len = max(cur_len, right_len + 1)
self.max_len = max(self.max_len, cur_len)
return cur_len
longestConsecutiveHelper(root)
return self.max_len