需要画图理解，递归是好写，但是性能垃圾
文章目录
- 前序遍历
- 中序遍历
- 后序遍历
前序遍历
思路：用栈实现，先处理当前节点，右孩子先入栈，左孩子后入栈。
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
res = []
if not root:
return res
stack = [root]
while stack:
node = stack.pop() # 先处理中间节点
res.append(node.val)
# 然后右孩子入栈，左孩子入栈
if node.right:
stack.append(node.right)
if node.left:
stack.append(node.left)
return res中序遍历
中序遍历不能和前序遍历一样，因为res存储的元素和访问的元素的顺序并非一致
中序遍历：左中右
前序遍历：中左右
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
res = []
if not root:
return res
stack = [] # 不能把node提前放进栈
cur = root
while cur or stack:
if cur:
stack.append(cur)
cur = cur.left
else:
node = stack.pop() # 中间节点
res.append(node.val)
cur = node.right
return res后序遍历
前序遍历是 中左右 左孩子先入栈 中右左 将结果翻转 得到左右中
后序遍历是 左右中
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
res = []
if not root:
return res
stack = [root]
while stack:
node = stack.pop()
res.append(node.val)
# 左孩子先入栈
if node.left:
stack.append(node.left)
if node.right:
stack.append(node.right)
return res[::-1]
