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# 二叉树的遍历
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!!! note "💡 一句话概述"
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二叉树遍历是按照特定规则走访每个节点恰好一次，核心分为深度优先（前/中/后序）和广度优先（层序）两大类。
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---
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## 🔑 核心概念
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1. **前序遍历（Pre-order）**：根 → 左 → 右，常用于序列化/复制树结构
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2. **中序遍历（In-order）**：左 → 根 → 右，BST 中序即有序序列
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3. **后序遍历（Post-order）**：左 → 右 → 根，常用于释放资源、计算目录大小
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4. **层序遍历（Level-order / BFS）**：逐层从左到右，借助队列实现
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---
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## 📝 详细说明
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### 遍历方式对比
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| 遍历方式 | 访问顺序 | 典型应用 | 实现方式 |
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|----------|----------|----------|----------|
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| 前序 | 根→左→右 | 序列化、复制树 | 递归 / 栈 |
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| 中序 | 左→根→右 | BST 排序输出 | 递归 / 栈 |
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| 后序 | 左→右→根 | 释放节点、表达式求值 | 递归 / 栈 |
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| 层序 | 逐层从左到右 | 最短路径、按层处理 | 队列 |
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### 递归 vs 迭代
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递归写法简洁直观，但存在栈溢出风险（树深度过大时）；迭代写法用显式栈/队列模拟调用栈，更安全也更可控。面试中通常要求两种都能写。
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### 统一迭代模板
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三种 DFS 遍历的迭代写法可以用「标记法」统一：入栈时按相反顺序压入节点和 null 标记，遇到 null 时才真正访问节点，通过调整压入顺序即可切换前/中/后序。
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---
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## 💻 代码示例
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### 节点定义
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```python
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class TreeNode:
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def __init__(self, val=0, left=None, right=None):
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self.val = val
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self.left = left
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self.right = right
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```
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### 递归写法
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```python
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def preorder(root):
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if not root:
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return []
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return [root.val] + preorder(root.left) + preorder(root.right)
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def inorder(root):
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if not root:
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return []
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return inorder(root.left) + [root.val] + inorder(root.right)
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def postorder(root):
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if not root:
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return []
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return postorder(root.left) + postorder(root.right) + [root.val]
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```
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### 迭代写法
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```python
|
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def preorder_iter(root):
|
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"""前序迭代：栈，先右后左压入"""
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if not root:
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return []
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stack, res = [root], []
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while stack:
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node = stack.pop()
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res.append(node.val)
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if node.right:
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stack.append(node.right)
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if node.left:
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stack.append(node.left)
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return res
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def inorder_iter(root):
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"""中序迭代：一路向左入栈，弹出后转向右子树"""
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stack, res = [], []
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cur = root
|
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while cur or stack:
|
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while cur:
|
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stack.append(cur)
|
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cur = cur.left
|
||||
cur = stack.pop()
|
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res.append(cur.val)
|
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cur = cur.right
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return res
|
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def postorder_iter(root):
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"""后序迭代：前序(根左右)翻转 → 根右左 → 反转得左右根"""
|
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if not root:
|
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return []
|
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stack, res = [root], []
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while stack:
|
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node = stack.pop()
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res.append(node.val)
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if node.left:
|
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stack.append(node.left)
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if node.right:
|
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stack.append(node.right)
|
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return res[::-1]
|
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```
|
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|
||||
### 层序遍历（BFS）
|
||||
|
||||
```python
|
||||
from collections import deque
|
||||
|
||||
def levelorder(root):
|
||||
if not root:
|
||||
return []
|
||||
q, res = deque([root]), []
|
||||
while q:
|
||||
level = []
|
||||
for _ in range(len(q)):
|
||||
node = q.popleft()
|
||||
level.append(node.val)
|
||||
if node.left:
|
||||
q.append(node.left)
|
||||
if node.right:
|
||||
q.append(node.right)
|
||||
res.append(level)
|
||||
return res
|
||||
```
|
||||
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||||
### 统一迭代模板（标记法）
|
||||
|
||||
```python
|
||||
def inorder_unified(root):
|
||||
"""统一模板：切换压入顺序即可实现前/中/后序"""
|
||||
if not root:
|
||||
return []
|
||||
stack, res = [root], []
|
||||
while stack:
|
||||
node = stack.pop()
|
||||
if node is None:
|
||||
# null 标记，下一个就是真正要访问的节点
|
||||
val_node = stack.pop()
|
||||
res.append(val_node.val)
|
||||
else:
|
||||
# 中序：左 → 根 → 右，入栈顺序相反：右 → 根(含标记) → 左
|
||||
if node.right:
|
||||
stack.append(node.right)
|
||||
stack.append(node) # 节点本身
|
||||
stack.append(None) # null 标记
|
||||
if node.left:
|
||||
stack.append(node.left)
|
||||
return res
|
||||
```
|
||||
|
||||
---
|
||||
|
||||
## ⚠️ 常见陷阱
|
||||
|
||||
!!! warning "中序迭代漏掉右子树"
|
||||
在中序迭代中，弹出节点后必须 `cur = node.right`，而不是继续向左走。忘记转向右子树会导致死循环或遗漏节点。
|
||||
|
||||
!!! warning "后序迭代的反转技巧"
|
||||
后序迭代用「前序翻转法」时，注意压栈顺序是**先左后右**（与标准前序相反），这样弹出顺序变成 根→右→左，反转后才是 左→右→根。
|
||||
|
||||
!!! warning "层序遍历中 queue 长度要在循环外快照"
|
||||
`for _ in range(len(q))` 必须在 for 循环开始前锁住当前层的节点数。如果在循环内动态取 `len(q)`，会把新入队的下一层节点也算进来。
|
||||
|
||||
---
|
||||
|
||||
## 🏋️ 练习题
|
||||
|
||||
??? question "练习 1：LeetCode 144 — 二叉树的前序遍历"
|
||||
给你二叉树的根节点 `root`，返回它节点值的**前序遍历**。请用迭代方式实现。
|
||||
|
||||
??? success "答案"
|
||||
```python
|
||||
def preorderTraversal(root):
|
||||
if not root:
|
||||
return []
|
||||
stack, res = [root], []
|
||||
while stack:
|
||||
node = stack.pop()
|
||||
res.append(node.val)
|
||||
if node.right:
|
||||
stack.append(node.right)
|
||||
if node.left:
|
||||
stack.append(node.left)
|
||||
return res
|
||||
```
|
||||
|
||||
??? question "练习 2：LeetCode 94 — 二叉树的中序遍历"
|
||||
给定一个二叉树的根节点 `root`，返回它的**中序遍历**。请用迭代方式实现。
|
||||
|
||||
??? success "答案"
|
||||
```python
|
||||
def inorderTraversal(root):
|
||||
stack, res = [], []
|
||||
cur = root
|
||||
while cur or stack:
|
||||
while cur:
|
||||
stack.append(cur)
|
||||
cur = cur.left
|
||||
cur = stack.pop()
|
||||
res.append(cur.val)
|
||||
cur = cur.right
|
||||
return res
|
||||
```
|
||||
|
||||
??? question "练习 3：LeetCode 102 — 二叉树的层序遍历"
|
||||
给你二叉树的根节点 `root`，返回其节点值的**层序遍历**（即逐层地，从左到右访问所有节点）。
|
||||
|
||||
??? success "答案"
|
||||
```python
|
||||
from collections import deque
|
||||
|
||||
def levelOrder(root):
|
||||
if not root:
|
||||
return []
|
||||
q, res = deque([root]), []
|
||||
while q:
|
||||
level = []
|
||||
for _ in range(len(q)):
|
||||
node = q.popleft()
|
||||
level.append(node.val)
|
||||
if node.left:
|
||||
q.append(node.left)
|
||||
if node.right:
|
||||
q.append(node.right)
|
||||
res.append(level)
|
||||
return res
|
||||
```
|
||||
|
||||
---
|
||||
|
||||
## 🔗 相关链接
|
||||
|
||||
- [LeetCode 144. 二叉树的前序遍历](https://leetcode.cn/problems/binary-tree-preorder-traversal/) — 前序迭代练习
|
||||
- [LeetCode 94. 二叉树的中序遍历](https://leetcode.cn/problems/binary-tree-inorder-traversal/) — 中序迭代练习
|
||||
- [LeetCode 145. 二叉树的后序遍历](https://leetcode.cn/problems/binary-tree-postorder-traversal/) — 后序迭代练习
|
||||
- [LeetCode 102. 二叉树的层序遍历](https://leetcode.cn/problems/binary-tree-level-order-traversal/) — 层序 BFS 练习
|
||||
@@ -100,3 +100,4 @@ nav:
|
||||
- 计数布隆过滤器: algorithm/counting-bloom-filter.md
|
||||
- 布谷鸟过滤器: algorithm/cuckoo-filter.md
|
||||
- HeavyKeeper: algorithm/heavykeeper.md
|
||||
- 二叉树的遍历: algorithm/binary-tree-traversal.md
|
||||
|
||||
Reference in New Issue
Block a user