2024-10-24 Daily Challenge

Today I have done leetcode's October LeetCoding Challenge with cpp.

October LeetCoding Challenge 24

Description

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.

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
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

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