# [111. 二叉树的最小深度](https://leetcode.cn/problems/minimum-depth-of-binary-tree)
[English Version](/solution/0100-0199/0111.Minimum%20Depth%20of%20Binary%20Tree/README_EN.md)
## 题目描述
<!-- 这里写题目描述 -->
<p>给定一个二叉树，找出其最小深度。</p>
<p>最小深度是从根节点到最近叶子节点的最短路径上的节点数量。</p>
<p><strong>说明：</strong>叶子节点是指没有子节点的节点。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0111.Minimum%20Depth%20of%20Binary%20Tree/images/ex_depth.jpg" style="width: 432px; height: 302px;" />
<pre>
<strong>输入：</strong>root = [3,9,20,null,null,15,7]
<strong>输出：</strong>2
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>root = [2,null,3,null,4,null,5,null,6]
<strong>输出：</strong>5
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li>树中节点数的范围在 <code>[0, 10<sup>5</sup>]</code> 内</li>
<li><code>-1000 <= Node.val <= 1000</code></li>
</ul>
## 解法
<!-- 这里可写通用的实现逻辑 -->
**方法一：递归**
递归的终止条件是当前节点为空，此时返回 $0$；如果当前节点左右子树有一个为空，返回不为空的子树的最小深度加 $1$；如果当前节点左右子树都不为空，返回左右子树最小深度的较小值加 $1$。
时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点个数。
**方法二：BFS**
使用队列实现广度优先搜索，初始时将根节点加入队列。每次从队列中取出一个节点，如果该节点是叶子节点，则直接返回当前深度；如果该节点不是叶子节点，则将该节点的所有非空子节点加入队列。继续搜索下一层节点，直到找到叶子节点。
时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点个数。
<!-- tabs:start -->
### **Python3**
<!-- 这里可写当前语言的特殊实现逻辑 -->
```python
# 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 minDepth(self, root: Optional[TreeNode]) -> int:
if root is None:
return 0
if root.left is None:
return 1 + self.minDepth(root.right)
if root.right is None:
return 1 + self.minDepth(root.left)
return 1 + min(self.minDepth(root.left), self.minDepth(root.right))
```
```python
# 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 minDepth(self, root: Optional[TreeNode]) -> int:
if root is None:
return 0
q = deque([root])
ans = 0
while 1:
ans += 1
for _ in range(len(q)):
node = q.popleft()
if node.left is None and node.right is None:
return ans
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
```
### **Java**
<!-- 这里可写当前语言的特殊实现逻辑 -->
```java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int minDepth(TreeNode root) {
if (root == null) {
return 0;
}
if (root.left == null) {
return 1 + minDepth(root.right);
}
if (root.right == null) {
return 1 + minDepth(root.left);
}
return 1 + Math.min(minDepth(root.left), minDepth(root.right));
}
}
```
```java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int minDepth(TreeNode root) {
if (root == null) {
return 0;
}
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
int ans = 0;
while (true) {
++ans;
for (int n = q.size(); n > 0; n--) {
TreeNode node = q.poll();
if (node.left == null && node.right == null) {
return ans;
}
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
}
}
}
```
### **C++**
```cpp
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int minDepth(TreeNode* root) {
if (!root) {
return 0;
}
if (!root->left) {
return 1 + minDepth(root->right);
}
if (!root->right) {
return 1 + minDepth(root->left);
}
return 1 + min(minDepth(root->left), minDepth(root->right));
}
};
```
```cpp
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int minDepth(TreeNode* root) {
if (!root) {
return 0;
}
queue<TreeNode*> q{{root}};
int ans = 0;
while (1) {
++ans;
for (int n = q.size(); n; --n) {
auto node = q.front();
q.pop();
if (!node->left && !node->right) {
return ans;
}
if (node->left) {
q.push(node->left);
}
if (node->right) {
q.push(node->right);
}
}
}
}
};
```
### **Go**
```go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func minDepth(root *TreeNode) int {
if root == nil {
return 0
}
if root.Left == nil {
return 1 + minDepth(root.Right)
}
if root.Right == nil {
return 1 + minDepth(root.Left)
}
return 1 + min(minDepth(root.Left), minDepth(root.Right))
}
```
```go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func minDepth(root *TreeNode) (ans int) {
if root == nil {
return 0
}
q := []*TreeNode{root}
for {
ans++
for n := len(q); n > 0; n-- {
node := q[0]
q = q[1:]
if node.Left == nil && node.Right == nil {
return
}
if node.Left != nil {
q = append(q, node.Left)
}
if node.Right != nil {
q = append(q, node.Right)
}
}
}
}
```
### **JavaScript**
```js
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
var minDepth = function (root) {
if (!root) {
return 0;
}
if (!root.left) {
return 1 + minDepth(root.right);
}
if (!root.right) {
return 1 + minDepth(root.left);
}
return 1 + Math.min(minDepth(root.left), minDepth(root.right));
};
```
```js
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
var minDepth = function (root) {
if (!root) {
return 0;
}
const q = [root];
let ans = 0;
while (1) {
++ans;
for (let n = q.length; n; --n) {
const node = q.shift();
if (!node.left && !node.right) {
return ans;
}
if (node.left) {
q.push(node.left);
}
if (node.right) {
q.push(node.right);
}
}
}
};
```
### **TypeScript**
```ts
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function minDepth(root: TreeNode | null): number {
if (root == null) {
return 0;
}
const { left, right } = root;
if (left == null) {
return 1 + minDepth(right);
}
if (right == null) {
return 1 + minDepth(left);
}
return 1 + Math.min(minDepth(left), minDepth(right));
}
```
```ts
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function minDepth(root: TreeNode | null): number {
if (!root) {
return 0;
}
const q = [root];
let ans = 0;
while (1) {
++ans;
for (let n = q.length; n; --n) {
const node = q.shift();
if (!node.left && !node.right) {
return ans;
}
if (node.left) {
q.push(node.left);
}
if (node.right) {
q.push(node.right);
}
}
}
}
```
### **Rust**
```rust
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
if root.is_none() {
return 0;
}
let node = root.as_ref().unwrap().borrow();
if node.left.is_none() {
return 1 + Self::dfs(&node.right);
}
if node.right.is_none() {
return 1 + Self::dfs(&node.left);
}
1 + Self::dfs(&node.left).min(Self::dfs(&node.right))
}
pub fn min_depth(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
Self::dfs(&root)
}
}
```
### **C**
```c
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
#define min(a, b) (((a) < (b)) ? (a) : (b))
int minDepth(struct TreeNode* root) {
if (!root) {
return 0;
}
if (!root->left) {
return 1 + minDepth(root->right);
}
if (!root->right) {
return 1 + minDepth(root->left);
}
int left = minDepth(root->left);
int right = minDepth(root->right);
return 1 + min(left, right);
}
```
### **...**
```
```
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