226. Flip binary tree DFS method

226. Flip binary tree

Give you the root node of a binary tree root , Flip this binary tree , And return its root node .

Example 1：

Input ：root = [4,2,7,1,3,6,9]
Output ：[4,7,2,9,6,3,1]

Example 2：

Input ：root = [2,1,3]
Output ：[2,3,1]

Example 3：

Input ：root = []
Output ：[]

This problem seems very complicated , It's actually very simple , Let's just do it according to the rules ：

/** * Definition for a binary tree node. * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */

void   dfs(struct TreeNode* root){
    
    if(root){
    
        struct TreeNode* t;
        t=root->left;
        root->left=root->right;
        root->right=t;
        dfs(root->left);
        dfs(root->right);

    }
}



struct TreeNode* invertTree(struct TreeNode* root){
    
    dfs(root);
    return root;

}