给定一棵二叉树,返回节点值的先序遍历结果(尽量使用非递归算法)。
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def preorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
# 4 star, 必须熟练
rs = []
self.preorder(root, rs)
return rs
def preorder(self, node, rs):
if node:
rs.append(node.val)
self.preorder(node.left, rs)
self.preorder(node.right, rs)
class Solution(object):
def preorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
# 6 star, 递归版很简单,迭代版没写出来
# 使用栈,先入右子树,再入左子树,迭代时后入先出,因此会先迭代左子树
if not root:
return []
stack = [root]
ret = []
while stack:
cur = stack.pop()
ret.append(cur.val)
if cur.right:
stack.append(cur.right)
if cur.left:
stack.append(cur.left)
return ret