Solution.cpp

/**
 * 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) {}
 * };
 */
class Solution {
public:
    bool flipEquiv(TreeNode* root1, TreeNode* root2) {
        return dfs(root1, root2);
    }

    bool dfs(TreeNode* root1, TreeNode* root2) {
        if (root1 == root2 || (!root1 && !root2)) return true;
        if (!root1 || !root2 || root1->val != root2->val) return false;
        return (dfs(root1->left, root2->left) && dfs(root1->right, root2->right)) || (dfs(root1->left, root2->right) && dfs(root1->right, root2->left));
    }
};