Sword refers to Offer27-mirror of binary tree-easy

Question link

Title description:

Please complete a function, input a binary tree, and the function outputs its mirror image.

For example, enter:

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

Mirror output:

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

Test case:

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

data range:

0 <= 节点个数 <= 1000

Problem-solving ideas:

1. dfs, depth first search.
2. The search order is the left subtree first, then the right subtree. Finally swap the left and right subtrees
3. The recursive exit is a non-leaf node, just return.

AC code (c++)

/**
 * 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:

    void dfs(TreeNode* root){
    
    
        if(root == NULL){
    
    
            return ;
        }
        dfs(root->left);
        dfs(root->right);
        TreeNode * tmp = root->left;
        root->left = root->right;
        root->right = tmp; 
    }
    TreeNode* mirrorTree(TreeNode* root) {
    
    
        dfs(root);
        return root;
    }
};