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.
Flipped Trees Diagram

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].

Solution in 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:
    int ChildSum(TreeNode *node) {
        if (!node) return 0;

        int sum=0;
        if (node->left) sum += node->left->val;
        if (node->right) sum += node->right->val;
        return sum;
    }

    int ChildVal(TreeNode *node) {
        if (node) return node->val;
        return 0;
    }

    bool flipEquiv(TreeNode* root1, TreeNode* root2) {
        if (!root1 && !root2) return true;
        if (ChildVal(root1) != ChildVal(root2)) return false;
        if (ChildSum(root1)!=ChildSum(root2)) return false;

        if (root1->left) {
            if (root2->left && ChildVal(root1->left)==ChildVal(root2->left))
                return flipEquiv(root1->left, root2->left) && flipEquiv(root1->right, root2->right);
            else
                return flipEquiv(root1->left, root2->right) && flipEquiv(root1->right, root2->left);
        }
        else if (root1->right) {
            if (root2->right && ChildVal(root1->right)==ChildVal(root2->right))
                return flipEquiv(root1->right, root2->right) && flipEquiv(root1->left, root2->left);
            else
                return flipEquiv(root1->right, root2->left) && flipEquiv(root1->left, root2->right);
        }
        return true;
    }
};