# [101. 对称二叉树](https://leetcode.cn/problems/symmetric-tree)

[English Version](/solution/0100-0199/0101.Symmetric%20Tree/README_EN.md)

## 题目描述

<!-- 这里写题目描述 -->

<p>给你一个二叉树的根节点 <code>root</code> ， 检查它是否轴对称。</p>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0101.Symmetric%20Tree/images/symtree1.jpg" style="width: 354px; height: 291px;" />
<pre>
<strong>输入：</strong>root = [1,2,2,3,4,4,3]
<strong>输出：</strong>true
</pre>

<p><strong>示例 2：</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0101.Symmetric%20Tree/images/symtree2.jpg" style="width: 308px; height: 258px;" />
<pre>
<strong>输入：</strong>root = [1,2,2,null,3,null,3]
<strong>输出：</strong>false
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li>树中节点数目在范围 <code>[1, 1000]</code> 内</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>

<p>&nbsp;</p>

<p><strong>进阶：</strong>你可以运用递归和迭代两种方法解决这个问题吗？</p>

## 解法

<!-- 这里可写通用的实现逻辑 -->

**方法一：递归**

我们设计一个函数 $dfs(root1, root2)$，用于判断两个二叉树是否对称。答案即为 $dfs(root, root)$。

函数 $dfs(root1, root2)$ 的逻辑如下：

-   如果 $root1$ 和 $root2$ 都为空，则两个二叉树对称，返回 `true`；
-   如果 $root1$ 和 $root2$ 中只有一个为空，或者 $root1.val \neq root2.val$，则两个二叉树不对称，返回 `false`；
-   否则，判断 $root1$ 的左子树和 $root2$ 的右子树是否对称，以及 $root1$ 的右子树和 $root2$ 的左子树是否对称，这里使用了递归。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数。

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### **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 isSymmetric(self, root: Optional[TreeNode]) -> bool:
        def dfs(root1, root2):
            if root1 is None and root2 is None:
                return True
            if root1 is None or root2 is None or root1.val != root2.val:
                return False
            return dfs(root1.left, root2.right) and dfs(root1.right, root2.left)

        return dfs(root, root)
```

### **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 boolean isSymmetric(TreeNode root) {
        return dfs(root, root);
    }

    private boolean dfs(TreeNode root1, TreeNode root2) {
        if (root1 == null && root2 == null) {
            return true;
        }
        if (root1 == null || root2 == null || root1.val != root2.val) {
            return false;
        }
        return dfs(root1.left, root2.right) && dfs(root1.right, root2.left);
    }
}
```

### **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:
    bool isSymmetric(TreeNode* root) {
        function<bool(TreeNode*, TreeNode*)> dfs = [&](TreeNode* root1, TreeNode* root2) -> bool {
            if (!root1 && !root2) return true;
            if (!root1 || !root2 || root1->val != root2->val) return false;
            return dfs(root1->left, root2->right) && dfs(root1->right, root2->left);
        };
        return dfs(root, root);
    }
};
```

### **Go**

```go
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isSymmetric(root *TreeNode) bool {
	var dfs func(*TreeNode, *TreeNode) bool
	dfs = func(root1, root2 *TreeNode) bool {
		if root1 == nil && root2 == nil {
			return true
		}
		if root1 == nil || root2 == nil || root1.Val != root2.Val {
			return false
		}
		return dfs(root1.Left, root2.Right) && dfs(root1.Right, root2.Left)
	}
	return dfs(root, root)
}
```

### **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)
 *     }
 * }
 */

const dfs = (root1: TreeNode | null, root2: TreeNode | null) => {
    if (root1 == root2) {
        return true;
    }
    if (root1 == null || root2 == null || root1.val != root2.val) {
        return false;
    }
    return dfs(root1.left, root2.right) && dfs(root1.right, root2.left);
};

function isSymmetric(root: TreeNode | null): boolean {
    return dfs(root.left, root.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(root1: &Option<Rc<RefCell<TreeNode>>>, root2: &Option<Rc<RefCell<TreeNode>>>) -> bool {
        if root1.is_none() && root2.is_none() {
            return true;
        }
        if root1.is_none() || root2.is_none() {
            return false;
        }
        let node1 = root1.as_ref().unwrap().borrow();
        let node2 = root2.as_ref().unwrap().borrow();
        node1.val == node2.val
            && Self::dfs(&node1.left, &node2.right)
            && Self::dfs(&node1.right, &node2.left)
    }

    pub fn is_symmetric(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
        let node = root.as_ref().unwrap().borrow();
        Self::dfs(&node.left, &node.right)
    }
}
```

```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;
use std::collections::VecDeque;
impl Solution {
    pub fn is_symmetric(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
        let root = root.unwrap();
        let mut node = root.as_ref().borrow_mut();
        let mut queue = VecDeque::new();
        queue.push_back([node.left.take(), node.right.take()]);
        while let Some([root1, root2]) = queue.pop_front() {
            if root1.is_none() && root2.is_none() {
                continue;
            }
            if root1.is_none() || root2.is_none() {
                return false;
            }
            if let (Some(node1), Some(node2)) = (root1, root2) {
                let mut node1 = node1.as_ref().borrow_mut();
                let mut node2 = node2.as_ref().borrow_mut();
                if node1.val != node2.val {
                    return false;
                }
                queue.push_back([node1.left.take(), node2.right.take()]);
                queue.push_back([node1.right.take(), node2.left.take()]);
            }
        }
        true
    }
}
```

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