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binary-tree-right-side-view.py
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binary-tree-right-side-view.py
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# Time: O(n)
# Space: O(h)
#
# Given a binary tree, imagine yourself standing on the right side of it,
# return the values of the nodes you can see ordered from top to bottom.
#
# For example:
# Given the following binary tree,
# 1 <---
# / \
# 2 3 <---
# \ \
# 5 4 <---
# You should return [1, 3, 4].
#
# 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):
# @param root, a tree node
# @return a list of integers
def rightSideView(self, root):
result = []
self.rightSideViewDFS(root, 1, result)
return result
def rightSideViewDFS(self, node, depth, result):
if not node:
return
if depth > len(result):
result.append(node.val)
self.rightSideViewDFS(node.right, depth+1, result)
self.rightSideViewDFS(node.left, depth+1, result)
# BFS solution
# Time: O(n)
# Space: O(n)
class Solution2(object):
# @param root, a tree node
# @return a list of integers
def rightSideView(self, root):
if root is None:
return []
result, current = [], [root]
while current:
next_level = []
for i, node in enumerate(current):
if node.left:
next_level.append(node.left)
if node.right:
next_level.append(node.right)
if i == len(current) - 1:
result.append(node.val)
current = next_level
return result