# Leetcode: Binary Tree Preorder Traversal

Binary Tree Preorder Traversal

Similar Problems:

Given a binary tree, return the preorder traversal of its nodes’ values.

```For example:
Given binary tree [1,null,2,3],
1
\
2
/
3
return [1,2,3].
```

Note: Recursive solution is trivial, could you do it iteratively?

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

```## Blog link: https://code.dennyzhang.com/binary-tree-preorder-traversal
# 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]
"""
res = []
stack = []
p = root
# Whatever I have pushed to stack, I have visited already
while p:
res.append(p.val)
stack.append(p)
p = p.left

while len(stack) != 0:
top_item = stack.pop()
# get right
if top_item.right:
p = top_item.right
while p:
res.append(p.val)
stack.append(p)
p = p.left
return res

def preorderTraversal_v1(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
res = []
self.preorderTraversalRec(root, res)
return res
def preorderTraversalRec(self, root, list_value):
"""
:type root: TreeNode
:rtype: List[int]
"""
if root is None:
return
list_value.append(root.val)
if root.left:
self.preorderTraversalRec(root.left, list_value)
if root.right:
self.preorderTraversalRec(root.right, list_value)
```

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