# 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]
"""
if root == None:
return []
stack = [root]
res = []
while len(stack) != 0:
curNode = stack[-1]
del stack[-1]
res.append(curNode.val)
if curNode.right:
stack.append(curNode.right)
if curNode.left:
stack.append(curNode.left)
return res
# 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]
"""
if root == None:
return []
self.res = []
self.dfs(root)
return self.res
def dfs(self, node):
if node == None:
return
self.res.append(node.val)
self.dfs(node.left)
self.dfs(node.right)
https://leetcode.com/problems/binary-tree-preorder-traversal/#/description
Given a binary tree, return the preorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3}
,
1
\
2
/
3
return [1,2,3]
.
Note: Recursive solution is trivial, could you do it iteratively?