951. Flip Equivalent Binary Trees

For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees.

A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations.

Write a function that determines whether two binary trees are flip equivalent.  The trees are given by root nodes root1 and root2.

Example 1:

Input: root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]
Output: true
Explanation: We flipped at nodes with values 1, 3, and 5.

Note:

1. Each tree will have at most 100 nodes.
2. Each value in each tree will be a unique integer in the range [0, 99].

class Solution {

public:

    bool flipEquiv(TreeNode* root1, TreeNode* root2) {

        if (root1 == NULL) {

            return root2 == NULL;

        }

        if (root2 == NULL) {

            return root1 == NULL;

        }

        if (root1 -> val != root2 -> val) {

            return false;

        }

        return (flipEquiv(root1 -> left, root2 -> left) && flipEquiv(root1 -> right, root2 -> right)) || (flipEquiv(root1 -> left, root2 -> right) && flipEquiv(root1 -> right, root2 -> left));

        

    }

};

===== BFS =====

bool flipEquiv(TreeNode* root1, TreeNode* root2) {
        
        queue<TreeNode*> q1; 
        queue<TreeNode*> q2; 
        
        if(root1) q1.push(root1); 
        if(root2) q2.push(root2); 
        
        while(!q1.empty() && !q2.empty()){
            TreeNode* cur1 = q1.front(); 
            TreeNode* cur2 = q2.front(); 
        
            //cout << cur1->val << " :" << cur2->val  <<endl ; 
            if(cur1->val != cur2->val) return false; 
            
            if(cur1->left) q1.push(cur1->left);
            if(cur1->right) q1.push(cur1->right);
                
            if(cur1->left && cur2->left && cur1->left->val == cur2->left->val)
                q2.push(cur2->left);
            else if(cur1->left && cur2->right && cur1->left->val == cur2->right->val)
                q2.push(cur2->right);
            
            if(cur1->right&& cur2->right && cur1->right->val == cur2->right->val)
                q2.push(cur2->right);
            else if(cur1->right && cur2->left && cur1->right->val == cur2->left->val)
                q2.push(cur2->left);    
        
            q1.pop();
            q2.pop();
        }
        
        if(!q1.empty() || !q2.empty()) return false; 
        else return true;           
            
    }