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binary tree level order traversal.py
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binary tree level order traversal.py
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'''
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
'''
'''
Method:
Queue O(n) time O(n) space
'''
class Solution(object):
def levelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
if not root:
return []
res = []
queue = [root]
while queue:
nextNodes = [] # nodes for next level
level = [] # current level vals
for node in queue:
level.append(node.val)
if node.left:
nextNodes.append(node.left)
if node.right:
nextNodes.append(node.right)
res.append(level)
queue = nextNodes
return res
'''
Method:
recursion:
'''
class Solution(object):
def levelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
self.res = []
self.helper(root, 0)
return self.res
def helper(self, root, depth):
if not root:
return
if len(self.res) == depth: # if the new depth reached, assign a space for this level
self.res.append([])
self.res[depth].append(root.val)
self.helper(root.left, depth + 1)
self.helper(root.right, depth + 1)
'''
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
'''
'''
Queue/Stack O(n) time O(n) space
'''
class Solution(object):
def zigzagLevelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
if not root:
return []
res = []
stack = [root]
while stack:
nextNodes = [] # nodes for next level
level = [] # current level vals
for node in stack:
level.append(node.val)
if node.left:
nextNodes.append(node.left)
if node.right:
nextNodes.append(node.right)
if len(res) % 2 == 1: # reverse the level for zigzag
start = 0
end = len(level)
while start < end:
key = level[start]
level[start] = level[end]
level[end] = key
start += 1
end -= 1
res.append(level)
stack = nextNodes
return res
class Solution(object):
def zigzagLevelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
if not root:
return []
res = [[root.val]]
stack = [root]
while stack:
nextNodes = [] # nodes for next level
level = [] # current level vals
while stack:
node = stack.pop()
if len(res) % 2 == 0:
if node.left:
nextNodes.append(node.left)
level.append(node.left.val)
if node.right:
nextNodes.append(node.right)
level.append(node.right.val)
else:
if node.right:
nextNodes.append(node.right)
level.append(node.right.val)
if node.left:
nextNodes.append(node.left)
level.append(node.left.val)
if level:
res.append(level)
stack = nextNodes
return res
'''
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
'''
class Solution(object):
def levelOrderBottom(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
if not root:
return []
res = []
stack = [root]
while stack:
current = []
nextNodes = []
for node in stack:
current.append(node.val)
if node.left:
nextNodes.append(node.left)
if node.right:
nextNodes.append(node.right)
res = [current] + res
stack = nextNodes
return res