988. Flip Equivalent Binary Trees¶

LeetCode Problem View on GitHub

988. Flip Equivalent Binary Trees

Medium

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:

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:

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

Example 3:

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].

Solution¶

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean flipEquiv(TreeNode root1, TreeNode root2) {
        if (root1 == null && root2 == null) return true;
        if (root1 == null || root2 == null || root1.val != root2.val) return false;
        boolean current = flipEquiv(root1.left, root2.left) && flipEquiv(root1.right, root2.right);
        boolean flip = flipEquiv(root1.left, root2.right) && flipEquiv(root1.right, root2.left);
        return current || flip;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here