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:
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;
}
};