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Copy path144 Binary Tree Preorder Traversal.py
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144 Binary Tree Preorder Traversal.py
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"""
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? - see preTraverse_itr
Author: Rajeev Ranjan
"""
# 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):
"""Morris"""
ret = []
cur = root
while cur:
if not cur.left:
ret.append(cur.val)
cur = cur.right
else:
pre = cur.left
while pre.right and pre.right != cur:
pre = pre.right
if not pre.right:
pre.right = cur
ret.append(cur.val)
cur = cur.left
else:
pre.right = None
cur = cur.right
return ret
def preorderTraversal_memory(self, root):
"""
dfs
:param root:
:return:
"""
lst = []
self.preTraverse_itr(root, lst)
return lst
def preTraverse(self, node, lst):
if not node:
return
lst.append(node.val)
self.preTraverse(node.left, lst)
self.preTraverse(node.right, lst)
def preTraverse_itr(self, root, lst):
"""
stack = [R, L, cur]
Pretty simple using stack
double stacks
:param root:
:param lst:
:return:
"""
if not root:
return
stk = [root]
while stk:
node = stk.pop()
lst.append(node.val)
if node.right: # right first
stk.append(node.right)
if node.left:
stk.append(node.left)
if __name__=="__main__":
t1 = TreeNode(1)
t1.left = TreeNode(2)
print Solution().preorderTraversal(t1)