/**
 * 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 {
    int n,ans=1;
public:
    void dfs(TreeNode* root){
        if(!root) return;
        
 
if(!root->left&&!root->right){return;}
        if(root->left&&!root->right){dfs(root->left);return;}
        if(!root->left&&root->right){root->left=root->right;root->right=NULL;dfs(root->left);return;}
        if(root->left->val>root->right->val){
            swap(root
 
->left,root->right);
        }
        dfs(root->left);
        dfs(root->right);
    }
    void judge(TreeNode* root1,TreeNode* root2){
        if((root1==NULL&&root2!=NULL)||(root1!=NULL&&root2==NULL)){ans=0;return;}
        if(!root1) return;
        if(root1->val!=root2->val){ans=0
 
;return;}
        judge(root1->left,root2->left);
        judge(root1->right,root2->right);
    }
    bool flipEquiv(TreeNode* root1, TreeNode* root2) {
        //轉成左偏樹，
        n=0;
        dfs(root1);
        int a1=n;
        n=0;
        dfs(root2);
        if(a1!=n) return 0;
        judge(root1,root2);
        return ans;
    }
};

注意翻轉後進行單枝遞迴，並return