面试题 27. 二叉树的镜像

题目描述

请完成一个函数，输入一个二叉树，该函数输出它的镜像。

例如输入：

4 / \ 2 7 / \ / \ 1 3 6 9
镜像输出：

4 / \ 7 2 / \ / \ 9 6 3 1

示例 1：

输入：root = [4,2,7,1,3,6,9]
输出：[4,7,2,9,6,3,1]

限制：

0 <= 节点个数 <= 1000

注意：本题与主站 226 题相同：https://leetcode.cn/problems/invert-binary-tree/

解法

方法一：递归

我们先判断根节点是否为空，如果为空，直接返回空。如果不为空，我们交换根节点的左右子树，然后递归地交换左子树和右子树。

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。最坏情况下，二叉树退化为链表，递归深度为 $n$，因此系统使用 $O(n)$ 大小的栈空间。

Python3JavaC++GoTypeScriptRustJavaScriptC#Swift

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution:
    def mirrorTree(self, root: TreeNode) -> TreeNode:
        if root is None:
            return None
        root.left, root.right = root.right, root.left
        self.mirrorTree(root.left)
        self.mirrorTree(root.right)
        return root

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode mirrorTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        TreeNode t = root.left;
        root.left = root.right;
        root.right = t;
        mirrorTree(root.left);
        mirrorTree(root.right);
        return root;
    }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* mirrorTree(TreeNode* root) {
        if (!root) {
            return root;
        }
        swap(root->left, root->right);
        mirrorTree(root->left);
        mirrorTree(root->right);
        return root;
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func mirrorTree(root *TreeNode) *TreeNode {
    if root == nil {
        return root
    }
    root.Left, root.Right = root.Right, root.Left
    mirrorTree(root.Left)
    mirrorTree(root.Right)
    return root
}

/**
 * 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 mirrorTree(root: TreeNode | null): TreeNode | null {
    if (root == null) {
        return root;
    }
    const { left, right } = root;
    root.left = right;
    root.right = left;
    mirrorTree(left);
    mirrorTree(right);
    return root;
}

// 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::cell::RefCell;
use std::rc::Rc;
impl Solution {
    fn dfs(root: &mut Option<Rc<RefCell<TreeNode>>>) {
        if let Some(node) = root {
            let mut node = node.borrow_mut();
            let temp = node.left.take();
            node.left = node.right.take();
            node.right = temp;
            Self::dfs(&mut node.left);
            Self::dfs(&mut node.right);
        }
    }

    pub fn mirror_tree(mut root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {
        Self::dfs(&mut root);
        root
    }
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var mirrorTree = function (root) {
    if (!root) {
        return null;
    }
    const { left, right } = root;
    root.left = right;
    root.right = left;
    mirrorTree(left);
    mirrorTree(right);
    return root;
};

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode MirrorTree(TreeNode root) {
        if (root == null) {
            return root;
        }
        TreeNode t = root.left;
        root.left = root.right;
        root.right = t;
        MirrorTree(root.left);
        MirrorTree(root.right);
        return root;
    }
}

/* public class TreeNode {
*     var val: Int
*     var left: TreeNode?
*     var right: TreeNode?
*     init(_ val: Int) {
*         self.val = val
*         self.left = nil
*         self.right = nil
*     }
* }
*/

class Solution {
    func mirrorTree(_ root: TreeNode?) -> TreeNode? {
        guard let root = root else {
            return nil
        }
        let temp = root.left
        root.left = root.right
        root.right = temp
        _ = mirrorTree(root.left)
        _ = mirrorTree(root.right)
        return root
    }
}

面试题 27. 二叉树的镜像

题目描述

解法

方法一：递归

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