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★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ Consider all the leaves of a binary tree. From left to right order, the values of those leaves form a leaf value sequence. For example, in the given tree above, the leaf value sequence is Two binary trees are considered leaf-similar if their leaf value sequence is the same. Return Note:
请考虑一颗二叉树上所有的叶子,这些叶子的值按从左到右的顺序排列形成一个 叶值序列 。 举个例子,如上图所示,给定一颗叶值序列为 如果有两颗二叉树的叶值序列是相同,那么我们就认为它们是 叶相似 的。 如果给定的两个头结点分别为 提示:
Runtime: 12 ms
Memory Usage: 19 MB
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 leafSimilar(_ root1: TreeNode?, _ root2: TreeNode?) -> Bool { 16 return traverse(root1) == traverse(root2) 17 } 18 19 func traverse(_ root: TreeNode?) ->String 20 { 21 if root == nil {return String()} 22 if root?.left == nil && root?.right == nil 23 { 24 return String(root!.val) + "-" 25 } 26 return traverse(root?.left) + traverse(root?.right) 27 } 28 } 12ms
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 leafSimilar(_ root1: TreeNode?, _ root2: TreeNode?) -> Bool { 16 var leaves1 : [Int] = [] 17 var leaves2 : [Int] = [] 18 getLeaves(root1, &leaves1) 19 getLeaves(root2, &leaves2) 20 21 return isSame(leaves1, leaves2) 22 } 23 24 func getLeaves(_ root: TreeNode?, _ leaves: inout [Int]) { 25 guard let node = root else { 26 return 27 } 28 if node.left == nil && node.right == nil { 29 leaves.append(node.val) 30 return 31 } 32 getLeaves(node.left, &leaves) 33 getLeaves(node.right, &leaves) 34 } 35 36 func isSame(_ leaves1: [Int], _ leaves2: [Int]) -> Bool { 37 if leaves1.count != leaves2.count { 38 return false 39 } 40 41 for i in 0..<leaves1.count { 42 if leaves1[i] != leaves2[i] { 43 return false 44 } 45 } 46 return true 47 } 48 } Runtime: 16 ms Memory Usage: 19.1 MB
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 leafSimilar(_ root1: TreeNode?, _ root2: TreeNode?) -> Bool { 16 if root1 == nil { 17 if root2 == nil { 18 return true 19 } 20 return false 21 } 22 23 var roots1: [Int] = [], roots2: [Int] = [] 24 25 traversingBTree(root1!, roots: &roots1) 26 traversingBTree(root2!, roots: &roots2) 27 28 // 遍历并且存储叶节点的值至数组 29 30 guard roots1.count == roots2.count else { 31 return false 32 } 33 34 for index in 0..<roots1.count { 35 if roots1[index] != roots2[index] { 36 return false 37 } 38 } 39 40 return true 41 } 42 43 func traversingBTree(_ root: TreeNode, roots: inout [Int]) { 44 45 if root.left == nil && root.right == nil { 46 roots.append(root.val) 47 return ; 48 } 49 if let left = root.left { 50 traversingBTree(left, roots: &roots) 51 } 52 53 if let right = root.right { 54 traversingBTree(right, roots: &roots) 55 } 56 } 57 } 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 leafSimilar(_ root1: TreeNode?, _ root2: TreeNode?) -> Bool { 16 17 return getLeaves(root1) == getLeaves(root2) 18 19 } 20 func getLeaves(_ root:TreeNode?) -> [Int] 21 { 22 guard let root = root else 23 { 24 return [] 25 } 26 var result: [Int] = [] 27 if root.left == nil && root.right == nil 28 { 29 result += [root.val] 30 } 31 result += getLeaves(root.left) 32 result += getLeaves(root.right) 33 return result 34 } 35 } 24ms 1 class Solution { 2 func leafSimilar(_ root1: TreeNode?, _ root2: TreeNode?) -> Bool { 3 var arr1 = [Int]() 4 var arr2 = [Int]() 5 helper(root1, &arr1) 6 helper(root2, &arr2) 7 for i in 0..<min(arr1.count, arr2.count) { 8 if arr1[i] != arr2[i] { 9 return false 10 } 11 } 12 return true 13 } 14 15 func helper(_ root: TreeNode?, _ arr: inout [Int]) { 16 guard root?.left != nil || root?.right != nil else { 17 if root != nil { 18 arr.append(root!.val) 19 } 20 return 21 } 22 helper(root?.left, &arr) 23 helper(root?.right, &arr) 24 } 25 } 28ms 1 class Solution { 2 func leafSimilar(_ root1: TreeNode?, _ root2: TreeNode?) -> Bool { 3 var arr1 = [Int]() 4 var arr2 = [Int]() 5 6 getArray(root1, &arr1) 7 getArray(root2, &arr2) 8 9 return arr1 == arr2 10 } 11 func getArray(_ root: TreeNode?, _ arr: inout [Int]) { 12 guard let r = root else { return } 13 if r.left == nil && r.right == nil { 14 arr.append(r.val) 15 } else { 16 getArray(r.left, &arr) 17 getArray(r.right, &arr) 18 } 19 } 20 } 32ms 1 class Solution { 2 func leafSimilar(_ root1: TreeNode?, _ root2: TreeNode?) -> Bool { 3 var array1: [Int] = [] 4 var array2: [Int] = [] 5 dfs(root: root1, result: &array1) 6 dfs(root: root2, result: &array2) 7 return array1 == array2 8 } 9 10 func dfs(root: TreeNode?, result: inout [Int]) -> Void { 11 if let r = root { 12 if r.left == nil && r.right == nil { 13 result.append(r.val) 14 } 15 dfs(root: r.left, result: &result) 16 dfs(root: r.right, result: &result) 17 } 18 } 19 }
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