144. 二叉树的前序遍历
给定一个二叉树，返回它的 前序 遍历。
示例:
输入: [1,null,2,3]
1
2
/
3
输出: [1,2,3]
进阶: 递归算法很简单，你可以通过迭代算法完成吗？
递归
# 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: TreeNode) -> List[int]:
res=[]
def helper(root):
if not root :
return
res.append(root.val)
if root.left:
helper(root.left)
if root.right:
helper(root.right)
helper(root)
return res
迭代
# 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: TreeNode) -> List[int]:
if not root:
return []
stack_=[root]
res=[]
while stack_:
temp=stack_.pop()
res.append(temp.val)
if temp.right:
stack_.append(temp.right)
if temp.left:
stack_.append(temp.left)
return res
# 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: TreeNode) -> List[int]:
res=[]
stack=[]
cur=root
while cur or stack :
while cur:
res.append(cur.val)
stack.append(cur)
cur=cur.left
cur=stack.pop()
cur=cur.right
return res
94. 二叉树的中序遍历
给定一个二叉树，返回它的中序 遍历。
示例:
输入: [1,null,2,3]
1
2
/
3
输出: [1,3,2]
进阶: 递归算法很简单，你可以通过迭代算法完成吗？
递归
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
res=[]
def helper (root):
if not root:
return
helper(root.left)
res.append(root.val)
helper(root.right)
helper(root)
return res
迭代
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
res=[]
stack=[(0,root)]
while stack:
color,node=stack.pop()
if not node :
continue
if color==0:
stack.append((0,node.right))
stack.append((1,node))
stack.append((0,node.left))
else :
res.append(node.val)
return res
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def inorderTraversal(self, root: TreeNode) -> List[int]:
res=[]
stack=[]
cur=root
while stack or cur:
while cur:
stack.append(cur)
cur=cur.left
cur=stack.pop()
res.append(cur.val)
cur=cur.right
return res
145. 二叉树的后序遍历
给定一个二叉树，返回它的 后序 遍历。
示例:
输入: [1,null,2,3]
1
2
/
3
输出: [3,2,1]
进阶: 递归算法很简单，你可以通过迭代算法完成吗？
递归
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
res=[]
def helper(root):
if not root:
return
helper(root.left)
helper(root.right)
res.append(root.val)
helper(root)
return res
迭代
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
res=[]
stack=[root]
while stack:
cur=stack.pop()
if not cur:
continue
res.append(cur.val)
if cur.left:
stack.append(cur.left)
if cur.right:
stack.append(cur.right)
return res[::-1]
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
res=[]
stack=[]
cur=root
while stack or cur:
while cur:
stack.append(cur)
cur=cur.left if cur.left else cur.right
cur=stack.pop()
res.append(cur.val)
if stack and stack[-1].left==cur:
cur=stack[-1].right
else:
cur=None
return res
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
res=[]
stack=[(0,root)]
while stack:
color,node=stack.pop()
if not node :
continue
if color==0:
stack.append((1,node))
stack.append((0,node.right))
stack.append((0,node.left))
else :
res.append(node.val)
return res
102. 二叉树的层序遍历
给你一个二叉树，请你返回其按 层序遍历 得到的节点值。 （即逐层地，从左到右访问所有节点）。
示例：
二叉树：[3,9,20,null,null,15,7],
3
/
9 20
/
15 7
返回其层次遍历结果：
[
[3],
[9,20],
[15,7]
]
通过次数111,934提交次数181,382
在真实的面试中遇到过这道题？
递归
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
res=[]
def helper (root,depth):
if not root:
return
if len(res)<=depth:
res.append([])
res[depth].append(root.val)
helper(root.left,depth+1)
helper(root.right,depth+1)
helper(root,0)
return res
迭代
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
res=[]
queue=[root]
while queue:
temp1=queue[:]
queue=[]
temp2=[]
while temp1:
cur=temp1.pop(0)
if not cur:
continue
temp2.append(cur.val)
queue.append(cur.left)
queue.append(cur.right)
if temp2:
res.append(temp2)
return res
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
if root is None:
return []
q = [root]
result = []
while q:
res=[]
for i in range(len(q)):
cur=q.pop(0)
res.append(cur.val)
if cur.left:
q.append(cur.left)
if cur.right:
q.append(cur.right)
result.append(res)
return result
有一点时间没刷数据结构的题了，感觉都有点生疏了，趁着假期的尾巴把二叉树的遍历整理一遍。感觉还不错。