951. Flip Equivalent Binary Trees

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951. Flip Equivalent Binary Trees

Description

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.

Given the roots of two binary trees root1 and root2, return true if the two trees are flip equivalent or false otherwise.

Example 1:
Flipped Trees Diagram

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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.

Example 2:

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Input: root1 = [], root2 = []
Output: true

Example 3:

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Input: root1 = [], root2 = [1]
Output: false

Constraints:

  • The number of nodes in each tree is in the range [0, 100].
  • Each tree will have unique node values in the range [0, 99].

Hints/Notes

  • N/A

Solution

Language: C++

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/**
 * 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) {
        if (!root1 || !root2) {
            return root1 == root2;
        }

        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));
    }
};