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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:

1. Each tree will have at most 100 nodes.
2. Each value in each tree will be a unique integer in the range [0, 99].

Related Topics:
Tree

## Solution 1.

• /**
* 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;
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();
}
}

############

/**
* 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) {
return dfs(root1, root2);
}

private boolean dfs(TreeNode root1, TreeNode root2) {
if (root1 == root2 || (root1 == null && root2 == null)) {
return true;
}
if (root1 == null || root2 == null || root1.val != root2.val) {
return false;
}
return (dfs(root1.left, root2.left) && dfs(root1.right, root2.right))
|| (dfs(root1.left, root2.right) && dfs(root1.right, root2.left));
}
}

• // 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:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def flipEquiv(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
def dfs(root1, root2):
if root1 == root2 or (root1 is None and root2 is None):
return True
if root1 is None or root2 is None or root1.val != root2.val:
return False
return (dfs(root1.left, root2.left) and dfs(root1.right, root2.right)) or (
dfs(root1.left, root2.right) and dfs(root1.right, root2.left)
)

return dfs(root1, root2)

############

# 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))

• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func flipEquiv(root1 *TreeNode, root2 *TreeNode) bool {
var dfs func(root1, root2 *TreeNode) bool
dfs = func(root1, root2 *TreeNode) bool {
if root1 == root2 || (root1 == nil && root2 == nil) {
return true
}
if root1 == nil || root2 == nil || root1.Val != root2.Val {
return false
}
return (dfs(root1.Left, root2.Left) && dfs(root1.Right, root2.Right)) || (dfs(root1.Left, root2.Right) && dfs(root1.Right, root2.Left))
}
return dfs(root1, root2)
}