LeetCode（算法）- 226. 翻转二叉树

Posted 2022-04-23 放羊的牧码

题目链接：点击打开链接

题目大意：略

解题思路：略

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AC 代码

• 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;
 *     
 * 
 */

// 解决方案(1)
class Solution 
    public TreeNode invertTree(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 invertTree(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 invertTree(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() : 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) 
 * ;
 */

// 解决方案(1)
class Solution 
public:
    TreeNode* invertTree(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* invertTree(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;
    
;