[LeetCode] 144. Binary Tree Preorder Traversal 二叉树的先序遍历
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?
树的遍历，最常见的有先序遍历，中序遍历，后序遍历和层序遍历，它们用递归实现起来都非常的简单。而题目的要求是不能使用递归求解。
1. 用迭代和stack。2. Morris Traversal Solution
Python: Stack, Time: O(n), Space: O(h) # h is the height of the tree
class Solution2(object):
def preorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
result, stack = [], [(root, False)]
while stack:
root, is_visited = stack.pop()
if root is None:
continue
if is_visited:
result.append(root.val)
else:
stack.append((root.right, False))
stack.append((root.left, False))
stack.append((root, True))
return result
Python: Morris， Time: O(n), Space: O(1)
class Solution(object):
def preorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
result, curr = [], root
while curr:
if curr.left is None:
result.append(curr.val)
curr = curr.right
else:
node = curr.left
while node.right and node.right != curr:
node = node.right
if node.right is None:
result.append(curr.val)
node.right = curr
curr = curr.left
else:
node.right = None
curr = curr.right
return result
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