Formatted question description: https://leetcode.ca/all/951.html
951. 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.
Write a function that determines whether two binary trees are flip equivalent. The trees are given by root nodes root1
and root2
.
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.![]()
Note:
- Each tree will have at most
100
nodes. - Each value in each tree will be a unique integer in the range
[0, 99]
.
Related Topics:
Tree
Solution 1.
// OJ: https://leetcode.com/problems/flip-equivalent-binary-trees/
// Time: O(N)
// Space: O(logN)
class Solution {
public:
bool flipEquiv(TreeNode* root1, TreeNode* root2) {
if (!root1 && !root2) return true;
if (!root1 || !root2) return false;
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));
}
};
Java
-
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public boolean flipEquiv(TreeNode root1, TreeNode root2) { if (root1 == null && root2 == null) return true; if (root1 == null ^ root2 == null) return false; Queue<TreeNode> queue1 = new LinkedList<TreeNode>(); Queue<TreeNode> queue2 = new LinkedList<TreeNode>(); queue1.offer(root1); queue2.offer(root2); while (!queue1.isEmpty()) { if (queue2.isEmpty()) return false; TreeNode node1 = queue1.poll(), node2 = queue2.poll(); if (node1.val != node2.val) return false; TreeNode left1 = node1.left, right1 = node1.right, left2 = node2.left, right2 = node2.right; if (left1 == null && right1 == null) { if (left2 != null || right2 != null) return false; } else { if (left1 == null) { left1 = right1; right1 = null; } if (left2 == null) { left2 = right2; right2 = null; } if (left1 == null ^ left2 == null) return false; if (right1 == null ^ right2 == null) return false; if (right1 == null && left1.val != left2.val) return false; if (left1 != null && right1 != null) { if (left1.val == right2.val) { TreeNode temp = left2; left2 = right2; right2 = temp; } } if (left1 != null) queue1.offer(left1); if (right1 != null) queue1.offer(right1); if (left2 != null) queue2.offer(left2); if (right2 != null) queue2.offer(right2); } } return queue2.isEmpty(); } }
-
// OJ: https://leetcode.com/problems/flip-equivalent-binary-trees/ // Time: O(N) // Space: O(logN) class Solution { public: bool flipEquiv(TreeNode* root1, TreeNode* root2) { if (!root1 && !root2) return true; if (!root1 || !root2) return false; 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)); } };
-
# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def flipEquiv(self, root1, root2): """ :type root1: TreeNode :type root2: TreeNode :rtype: bool """ if not root1 and not root2: return True if not root1 and root2: return False if root1 and not root2: return False if root1.val != root2.val: return False return (self.flipEquiv(root1.left, root2.right) and self.flipEquiv(root1.right, root2.left)) or (self.flipEquiv(root1.left, root2.left) and self.flipEquiv(root1.right, root2.right))