LeetCode 951: Flip Equivalent Binary Trees Solution
Master LeetCode problem 951 (Flip Equivalent Binary Trees), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
951. Flip Equivalent Binary Trees
Problem Explanation
Explanation
To determine if two binary trees are flip equivalent, we can recursively check the following conditions:
- If both roots are null, return true.
- If only one of the roots is null, return false.
- If the values of the roots are different, return false.
- Recursively check if the left subtrees are flip equivalent and if the right subtrees are flip equivalent, or swap the left and right children of one of the trees and check again.
Time complexity: O(min(N1, N2)), where N1 and N2 are the number of nodes in the two trees. Space complexity: O(min(H1, H2)), where H1 and H2 are the heights of the two trees.
Solution Code
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;
}
return (flipEquiv(root1.left, root2.left) && flipEquiv(root1.right, root2.right)) ||
(flipEquiv(root1.left, root2.right) && flipEquiv(root1.right, root2.left));
}
}Try It Yourself
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