LeetCode（剑指 Offer）- 27. 二叉树的镜像

题目链接：​​点击打开链接​​

题目大意：略

解题思路：略

相关企业

• 字节跳动
• 谷歌（Google）
• Facebook
• 亚马逊（Amazon）
• 优步（Uber）
• 微软（Microsoft）
• 苹果（Apple）

AC 代码

• Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 
// 解决方案(1)
class Solution {
    public TreeNode mirrorTree(TreeNode root) {
        dfs(root);
        return root;
    }
 
    private void dfs(TreeNode node) {
        if (null == node) {
            return;
        }
 
        dfs(node.left);
        dfs(node.right);
 
        TreeNode t = node.left;
        node.left = node.right;
        node.right = t;
    }
}
 
// 解决方案(2)
class Solution {
    public TreeNode mirrorTree(TreeNode root) {
        if(root == null) return null;
        Stack<TreeNode> stack = new Stack<>() {{ add(root); }};
        while(!stack.isEmpty()) {
            TreeNode node = stack.pop();
            if(node.left != null) stack.add(node.left);
            if(node.right != null) stack.add(node.right);
            TreeNode tmp = node.left;
            node.left = node.right;
            node.right = tmp;
        }
        return root;
    }
}
 
// 解决方案(3)
class Solution {
    public TreeNode mirrorTree(TreeNode root) {
        if(root == null) return null;
        TreeNode tmp = root.left;
        root.left = mirrorTree(root.right);
        root.right = mirrorTree(tmp);
        return root;
    }
}

• C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 
// 解决方案(1)
class Solution {
public:
    TreeNode* mirrorTree(TreeNode* root) {
        if (root == nullptr) return nullptr;
        TreeNode* tmp = root->left;
        root->left = mirrorTree(root->right);
        root->right = mirrorTree(tmp);
        return root;
    }
};
 
// 解决方案(2)
class Solution {
public:
    TreeNode* mirrorTree(TreeNode* root) {
        if(root == nullptr) return nullptr;
        stack<TreeNode*> stack;
        stack.push(root);
        while (!stack.empty())
        {
            TreeNode* node = stack.top();
            stack.pop();
            if (node->left != nullptr) stack.push(node->left);
            if (node->right != nullptr) stack.push(node->right);
            TreeNode* tmp = node->left;
            node->left = node->right;
            node->right = tmp;
        }
        return root;
    }
};