LeetCode解题之Binary Tree Inorder Traversal
原题
不用递归来实现树的中序遍历。
注意点:
- 无
例子:
输入: {1,#,2,3}
1
2
/
3
输出: [1,3,2]
解题思路
通过栈来实现,从根节点开始,不断寻找左节点,并把这些节点依次压入栈内,只有在该节点没有左节点或者它的左子树都已经遍历完成后,它才会从栈内弹出,这时候访问该节点,并它的右节点当做新的根节点一样不断遍历。
AC源码
# 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 inorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
result = []
stack = []
p = root
while p or stack:
# Save the nodes which have left child
while p:
stack.append(p)
p = p.left
if stack:
p = stack.pop()
# Visit the middle node
result.append(p.val)
# Visit the right subtree
p = p.right
return result
if __name__ == "__main__":
n1 = TreeNode(1)
n2 = TreeNode(2)
n3 = TreeNode(3)
n1.right = n2
n2.left = n3
assert Solution().inorderTraversal(n1) == [1, 3, 2]
欢迎查看我的Github (https://github.com/gavinfish/LeetCode-Python) 来获得相关源码。
时间: 2024-10-10 12:32:28