Skip to content

951. Flip Equivalent Binary Trees

Description

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].

Solutions

Solution 1

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
# 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)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
/**
 * 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));
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool flipEquiv(TreeNode* root1, TreeNode* root2) {
        return dfs(root1, root2);
    }

    bool dfs(TreeNode* root1, TreeNode* root2) {
        if (root1 == root2 || (!root1 && !root2)) return true;
        if (!root1 || !root2 || 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));
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
/**
 * 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)
}

1
2
3
4
5
6
7
8
9
function flipEquiv(root1: TreeNode | null, root2: TreeNode | null): boolean {
    if (root1 === root2) return true;
    if (!root1 || !root2 || root1?.val !== root2?.val) return false;

    const { left: l1, right: r1 } = root1!;
    const { left: l2, right: r2 } = root2!;

    return (flipEquiv(l1, l2) && flipEquiv(r1, r2)) || (flipEquiv(l1, r2) && flipEquiv(r1, l2));
}

1
2
3
4
5
6
7
8
9
function flipEquiv(root1, root2) {
    if (root1 === root2) return true;
    if (!root1 || !root2 || root1?.val !== root2?.val) return false;

    const { left: l1, right: r1 } = root1;
    const { left: l2, right: r2 } = root2;

    return (flipEquiv(l1, l2) && flipEquiv(r1, r2)) || (flipEquiv(l1, r2) && flipEquiv(r1, l2));
}

Comments