[Swift]LeetCode95. 不同的二叉搜索树 II | Unique Binary Search Trees II
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Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ... n.
Example:
Input: 3
Output:
[
[1,null,3,2],
[3,2,null,1],
[3,1,null,null,2],
[2,1,3],
[1,null,2,null,3]
]
Explanation:
The above output corresponds to the 5 unique BST's shown below:
1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
给定一个整数 n，生成所有由 1 ... n 为节点所组成的二叉搜索树。
示例:
输入: 3
输出:
[
[1,null,3,2],
[3,2,null,1],
[3,1,null,null,2],
[2,1,3],
[1,null,2,null,3]
]
解释:
以上的输出对应以下 5 种不同结构的二叉搜索树：
1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
36ms
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 generateTrees(_ n: Int) -> [TreeNode?] { 16 if n == 0{ 17 return [TreeNode?]() 18 } 19 20 return build(1, n) 21 } 22 23 func build(_ start : Int, _ end : Int) -> [TreeNode?]{ 24 var res = [TreeNode?]() 25 if(start > end){ 26 res.append(nil) 27 return res 28 } 29 if(start == end){ 30 var node = TreeNode(start) 31 res.append(node) 32 return res 33 } 34 35 for i in start...end { 36 var left = build(start, i-1) 37 var right = build(i+1, end) 38 for leftNode in left{ 39 for rightNode in right{ 40 var node = TreeNode(i) 41 node.left = leftNode 42 node.right = rightNode 43 res.append(node) 44 } 45 } 46 47 } 48 return res 49 } 50 }
40ms
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 generateTrees(_ n: Int) -> [TreeNode?] { 16 if n == 0 {return []} 17 return genTrees(start: 1, end: n) 18 } 19 func genTrees (start: Int, end: Int) -> [TreeNode?] { 20 var ret: [TreeNode?] = [] 21 if start > end { 22 ret.append(nil) 23 return ret 24 } else if start == end { 25 var node = TreeNode(start) 26 ret.append(node) 27 return ret 28 } 29 var left: [TreeNode?] = [] 30 var right: [TreeNode?] = [] 31 for i in start ... end { 32 left = genTrees(start: start, end: i - 1) 33 right = genTrees(start: i + 1, end: end) 34 for left_node in left { 35 for right_node in right { 36 let root = TreeNode(i) 37 root.left = left_node 38 root.right = right_node 39 ret.append(root) 40 } 41 } 42 } 43 return ret 44 } 45 }
60ms
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 generateTrees(_ n: Int) -> [TreeNode?] { 16 if n == 0 { return [TreeNode?]() } 17 if n == 1 { return [TreeNode(1)] } 18 var matrix = [[TreeNode?]]() 19 matrix.append([nil]) 20 matrix.append([TreeNode(1)]) 21 22 func helper(array: [Int]) -> [TreeNode?] { 23 if array.count == 0 { 24 return [nil] 25 } else if array.count == 1 { 26 return [TreeNode(array[0])] 27 } else { 28 var result = [TreeNode?]() 29 for i in 0 ..< array.count { 30 31 var temp = i == 0 ? [Int]() : Array(array[0...i - 1]) 32 let leftArray = helper(array: temp) 33 temp = i == array.count - 1 ? [Int]() : Array(array[i + 1...array.count - 1]) 34 let rightArray = helper(array: temp) 35 for left in leftArray { 36 for right in rightArray { 37 let root = TreeNode(array[i]) 38 root.left = left 39 root.right = right 40 result.append(root) 41 } 42 } 43 44 } 45 return result 46 } 47 } 48 49 return helper(array: Array(1 ... n)) 50 } 51 }
