Description
Given a binary tree, return the preorder traversal of its nodes’ values.
Example:
Input: [1,null,2,3]
1
\
2
/
3
Output: [1,2,3]
Solutions
1. 递归方法
# Time Complexity: O(logn)
# Space Complexity: O(1)
# 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 _preorderTraversal(self, root, res):
if root:
res.append(root.val)
self._preorderTraversal(root.left, res)
self._preorderTraversal(root.right, res)
def preorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
res = []
self._preorderTraversal(root, res)
return res
# Runtime: 16 ms, faster than 83.31% of Python online submissions for Binary Tree Preorder Traversal.
# Memory Usage: 11.9 MB, less than 9.67% of Python online submissions for Binary Tree Preorder Traversal.
2. DFS-迭代
三个遍历顺序基本上都是用 Stack 实现,后序可以按照先序的结果逆序就好。先序遍历时节点入栈的时候需要注意要右节点先,左节点后。
# Time Complextiy: O(logn)
# Space Complexity: O(logn)
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def preorderTraversal(self, root: TreeNode) -> List[int]:
stack = [root]
res = []
while stack:
node = stack.pop()
if not node:
continue
res.append(node.val)
stack.append(node.right) # right first push, last pop
stack.append(node.left)
return res
# Runtime: 40 ms, faster than 38.84% of Python3 online submissions for Binary Tree Preorder Traversal.
# Memory Usage: 13.9 MB, less than 6.52% of Python3 online submissions for Binary Tree Preorder Traversal.
迭代就会难一些了,这里不能像一般的 DFS 那样 stack 直接放到 while 后面做条件判断了,这里加上 node。
# Time Complexity: O(logn)
# Space Complexity: O(n)
# 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 preorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
stack = []
res = []
node = root
while node or stack:
if node:
res.append(node.val)
stack.append(node.right)
node = node.left
else:
node = stack.pop()
return res
# Runtime: 20 ms, faster than 61.96% of Python online submissions for Binary Tree Preorder Traversal.
# Memory Usage: 11.7 MB, less than 72.52% of Python online submissions for Binary Tree Preorder Traversal.
用循环找的方式:
# 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 preorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
stack = []
res = []
node = root
while node or stack:
while node:
res.append(node.val)
stack.append(node)
node = node.left
node = stack.pop()
node = node.right
return res
# Runtime: 16 ms, faster than 83.31% of Python online submissions for Binary Tree Preorder Traversal.
# Memory Usage: 11.8 MB, less than 54.63% of Python online submissions for Binary Tree Preorder Traversal.
# Next challenges: