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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
100nodes. - Each value in each tree will be a unique integer in the range
[0, 99].
我们可以为二叉树 T 定义一个翻转操作，如下所示：选择任意节点，然后交换它的左子树和右子树。
只要经过一定次数的翻转操作后，能使 X 等于 Y，我们就称二叉树 X 翻转等价于二叉树 Y。
编写一个判断两个二叉树是否是翻转等价的函数。这些树由根节点 root1和 root2 给出。
示例：
输入：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] 输出：true 解释：We flipped at nodes with values 1, 3, and 5.
提示：
- 每棵树最多有
100个节点。 - 每棵树中的每个值都是唯一的、在
[0, 99]范围内的整数。
16ms
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * public var val: Int 5 * public var left: TreeNode? 6 * public var right: TreeNode? 7 * public init(_ val: Int) { 8 * self.val = val 9 * self.left = nil 10 * self.right = nil 11 * } 12 * } 13 */ 14 class Solution { 15 func flipEquiv(_ root1: TreeNode?, _ root2: TreeNode?) -> Bool { 16 if root1 == nil {return root2 == nil} 17 if root2 == nil {return root1 == nil} 18 if root1!.val != root2!.val {return false} 19 if flipEquiv(root1!.left, root2!.left) && flipEquiv(root1!.right, root2!.right) 20 { 21 return true 22 } 23 if flipEquiv(root1!.left, root2!.right) && flipEquiv(root1!.right, root2!.left) 24 { 25 return true 26 } 27 return false 28 } 29 }
16ms
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * public var val: Int 5 * public var left: TreeNode? 6 * public var right: TreeNode? 7 * public init(_ val: Int) { 8 * self.val = val 9 * self.left = nil 10 * self.right = nil 11 * } 12 * } 13 */ 14 class Solution { 15 func flipEquiv(_ root1: TreeNode?, _ root2: TreeNode?) -> Bool { 16 if root1 == nil && root2 == nil { 17 return true 18 } 19 guard let root1 = root1, let root2 = root2 else { 20 return false 21 } 22 guard root1.val == root2.val else { 23 return false 24 } 25 return (flipEquiv(root1.left, root2.left) && flipEquiv(root1.right, root2.right)) || (flipEquiv(root1.left, root2.right) && flipEquiv(root1.right, root2.left)) 26 } 27 }
20ms
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * public var val: Int 5 * public var left: TreeNode? 6 * public var right: TreeNode? 7 * public init(_ val: Int) { 8 * self.val = val 9 * self.left = nil 10 * self.right = nil 11 * } 12 * } 13 */ 14 class Solution { 15 func flipEquiv(_ root1: TreeNode?, _ root2: TreeNode?) -> Bool { 16 if root1 == nil && root2 == nil { return true } 17 if (root1 != nil && root2 == nil) || (root1 == nil && root2 != nil) {return false} 18 19 var root1 = root1 20 21 var level1: [TreeNode?] = [root1] 22 var level2: [TreeNode?] = [root2] 23 var level1n: [TreeNode?] = [] 24 var level2n: [TreeNode?] = [] 25 var rr = 0 26 while level1.count > 0 { 27 if level1.count != level2.count { return false } 28 print(rr) 29 rr += 1 30 if level1.count == 1 { 31 if level1[0] == nil && level2[0] != nil { return false } 32 if level1[0] != nil && level2[0] == nil { return false } 33 if level1[0] == nil && level2[0] == nil { continue } 34 if level1[0]?.val != level2[0]?.val { return false } 35 } else { 36 var i = 0 37 while i < level1.count - 1 { 38 defer { i += 2 } 39 if level1[i] == nil && level2[i] == nil && level1[i+1] == nil && level2[i+1] == nil { continue } 40 if level1[i] != nil && level2[i] != nil && level1[i]?.val == level2[i]?.val && level1[i+1] != nil && level2[i+1] != nil && level1[i+1]?.val == level2[i+1]?.val { continue } 41 42 if level1[i] != nil && level2[i] != nil && level1[i+1] == nil && level2[i+1] == nil && level1[i]?.val == level2[i]?.val { continue } 43 44 if level1[i] == nil && level2[i] == nil && level1[i+1] != nil && level2[i+1] != nil && level1[i+1]?.val == level2[i+1]?.val { continue } 45 46 if level1[i] != nil && level2[i] != nil && level1[i+1] != nil && level2[i+1] != nil && level1[i]?.val == level2[i+1]?.val { 47 var tmp = level1[i] 48 level1[i] = level1[i+1] 49 level1[i+1] = tmp 50 continue 51 } 52 if level1[i] != nil && level2[i] == nil && level1[i+1] == nil && level2[i+1] != nil && level1[i]?.val == level2[i+1]?.val { 53 var tmp = level1[i] 54 level1[i] = level1[i+1] 55 level1[i+1] = tmp 56 continue 57 } 58 59 if level1[i] == nil && level2[i] != nil && level1[i+1] != nil && level2[i+1] == nil && level1[i+1]?.val == level2[i]?.val { 60 var tmp = level1[i] 61 level1[i] = level1[i+1] 62 level1[i+1] = tmp 63 continue 64 } 65 66 return false 67 } 68 } 69 70 for node in level1 { 71 if let node = node { 72 level1n.append(node.left) 73 level1n.append(node.right) 74 } 75 } 76 77 for node in level2 { 78 if let node = node { 79 level2n.append(node.left) 80 level2n.append(node.right) 81 } 82 } 83 84 level1 = level1n 85 level2 = level2n 86 level1n = [] 87 level2n = [] 88 } 89 90 return true 91 } 92 }
24ms
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * public var val: Int 5 * public var left: TreeNode? 6 * public var right: TreeNode? 7 * public init(_ val: Int) { 8 * self.val = val 9 * self.left = nil 10 * self.right = nil 11 * } 12 * } 13 */ 14 class Solution { 15 func flipEquiv(_ root1: TreeNode?, _ root2: TreeNode?) -> Bool { 16 17 if root1 == nil && root2 == nil { 18 return true 19 } else if root1 == nil { 20 return false 21 } else if root2 == nil { 22 return false 23 } 24 25 let _r1 = root1! 26 let _r2 = root2! 27 28 return _r1.val == _r2.val && 29 (flipEquiv(_r1.left, _r2.right) || flipEquiv(_r1.left, _r2.left)) && 30 (flipEquiv(_r1.right, _r2.left) || flipEquiv(_r1.right, _r2.right)) 31 } 32 }
40ms
1 class Solution { 2 static let size = 100 3 var m1: Array<Array<Int>> = [[Int]].init(repeating: [Int].init(repeating: -1, count: size), count: size) 4 var m2: Array<Array<Int>> = [[Int]].init(repeating: [Int].init(repeating: -1, count: size), count: size) 5 6 func flipEquiv(_ root1: TreeNode?, _ root2: TreeNode?) -> Bool { 7 var s1: Int = 0 8 var s2: Int = 0 9 treeMap(root1, &m1, 0, size: &s1) 10 treeMap(root2, &m2, 0, size: &s2) 11 if s1 != s2 { return false } 12 return checkEquiv(m1, m2) 13 } 14 15 func checkEquiv(_ r1: [[Int]], _ r2: [[Int]]) -> Bool { 16 for i in 0..<Solution.size { 17 for j in 0..<Solution.size { 18 if r1[i][j] != r2[i][j] { 19 return false 20 } 21 } 22 } 23 return true 24 } 25 26 func treeMap(_ node: TreeNode?, 27 _ arr: inout [[Int]], 28 _ lvl: Int, 29 size: inout Int) { 30 if let n = node { 31 size += 1 32 if let r = n.right { 33 arr[n.val][r.val] = lvl 34 treeMap(r, &arr, lvl+1, size: &size) 35 } 36 if let l = n.left { 37 arr[n.val][l.val] = lvl 38 treeMap(l, &arr, lvl+1, size: &size) 39 } 40 } 41 } 42 }