Given an integer array with no duplicates. A maximum tree building on this array is defined as follow:

1. The root is the maximum number in the array.
2. The left subtree is the maximum tree constructed from left part subarray divided by the maximum number.
3. The right subtree is the maximum tree constructed from right part subarray divided by the maximum number.

Construct the maximum tree by the given array and output the root node of this tree.

Example 1:

Input: [3,2,1,6,0,5]
Output: return the tree root node representing the following tree:

      6
    /      3     5
    \    / 
     2  0   
               1 

Note:

1. The size of the given array will be in the range [1,1000].

給定一個不含重復元素的整數數組。一個以此數組構建的最大二叉樹定義如下：

1. 二叉樹的根是數組中的最大元素。
2. 左子樹是通過數組中最大值左邊部分構造出的最大二叉樹。
3. 右子樹是通過數組中最大值右邊部分構造出的最大二叉樹。

通過給定的數組構建最大二叉樹，並且輸出這個樹的根節點。

Example 1:

輸入: [3,2,1,6,0,5]
輸入: 返回下面這棵樹的根節點：

      6
    /      3     5
    \    / 
     2  0   
               1

註意:

1. 給定的數組的大小在 [1, 1000] 之間。

104ms

 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 constructMaximumBinaryTree(_ nums: [Int]) -> TreeNode? {
16         let end = nums.count
17         if (end == 0) {
18             return nil
19         }
20         var nums = nums
21         return construct(&nums, 0, end-1)
22     }
23     
24     func construct(_ nums: inout [Int], _ start: Int, _ end: Int) -> TreeNode {
25         var maxNumber = Int.min; var maxIndex = 0
26 
27         for i in start...end {
28             if (nums[i] >= maxNumber) {
29                 maxNumber = nums[i]
30                 maxIndex = i
31             }
32         }
33         let head = TreeNode(maxNumber)
34         
35         if (start != end) {
36             switch maxIndex {
37                 case start:
38                 head.right = construct(&nums, start+1, end)
39                 case end:
40                 head.left = construct(&nums, start, end-1)
41                 default:
42                 head.left = construct(&nums, start, maxIndex-1)
43                 head.right = construct(&nums, maxIndex+1, end)
44             }
45         }
46         
47         return head
48     }
49 }

108ms

 1 class Solution {
 2     func constructMaximumBinaryTree(_ nums: [Int]) -> TreeNode? {        
 3         
 4         return construct(nums, 0, nums.count)
 5     }
 6     
 7     func construct(_ n: [Int], _ s: Int, _ e: Int) -> TreeNode? {
 8         if s == e { return nil }
 9         
10         let idx = max(n, s, e)
11         
12         var r = TreeNode(n[idx])
13         r.left = construct(n, s, idx)
14         r.right = construct(n, idx+1, e)
15         return r
16     }
17     
18     func max(_ nums: [Int], _ s: Int, _ e: Int) -> Int {
19         var idx = s
20         for i in s..<e {
21                 if nums[idx] < nums[i] { 
22                     idx = i 
23                 }
24             }
25         return idx
26     }
27 }

116ms

 1 class Solution {
 2     func constructMaximumBinaryTree(_ nums: [Int]) -> TreeNode? {
 3         return findMidMax(nums, 0, nums.count - 1)
 4     }
 5     
 6     func findMidMax(_ nums:[Int], _ i:Int, _ j:Int) -> TreeNode? {
 7         
 8         if i < 0 || i >= nums.count {
 9             return nil
10         } else if j < 0 || j >= nums.count {
11             return nil
12         } else if i > j {
13             return nil
14         }
15         
16         
17         
18         var maxNum = Int.min
19         var maxIndex = -1
20         var index = i
21         while (index <= j) {
22             if maxNum < nums[index]{
23                 maxNum = nums[index]
24                 maxIndex = index
25             }
26             index = index + 1
27         }
28         
29         var rootNode = TreeNode(nums[maxIndex])
30         rootNode.left = findMidMax(nums, i, maxIndex - 1)
31         rootNode.right = findMidMax(nums, maxIndex + 1, j)
32         
33         return rootNode
34     }
35 }

120ms

 1 class Solution {
 2     
 3     func constructMaximumBinaryTree(_ nums: [Int]) -> TreeNode? {
 4         var stack = [TreeNode]()
 5         
 6         for i in 0..<nums.count {
 7             var cur = TreeNode(nums[i])
 8             while let last = stack.last, last.val < nums[i] {
 9                 cur.left = last
10                 stack.popLast()
11             }
12             if (!stack.isEmpty) {
13                 stack.last?.right = cur
14             }
15             stack.append(cur)
16         }
17         return stack.first
18     }
19 }

132ms

 1 class Solution {
 2     func constructMaximumBinaryTree(_ nums: [Int]) -> TreeNode? {
 3         if nums.count == 0 { return nil }
 4         if nums.count == 1 { return TreeNode(nums[0]) }
 5         let maxIdx = getMaxIndex(nums)
 6         let tree = TreeNode(nums[maxIdx])
 7         tree.left = constructMaximumBinaryTree(Array(nums.prefix(maxIdx)))
 8         if nums.count > maxIdx {
 9             tree.right = constructMaximumBinaryTree(Array(nums.suffix(from: maxIdx+1)))
10         }
11         
12         return tree
13     }
14     
15     func getMaxIndex(_ arr: [Int]) -> Int {
16         var maxIdx = 0
17         for i in 0..<arr.count {
18             if arr[maxIdx] < arr[i] { maxIdx = i }
19         }
20         return maxIdx
21     }
22 }

 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 constructMaximumBinaryTree(_ nums: [Int]) -> TreeNode? {
16         var v:[TreeNode?] = [TreeNode?]()
17         for num in nums
18         {
19             var cur:TreeNode? = TreeNode(num)
20             while(!v.isEmpty && v.last!!.val < num)
21             {
22                 cur?.left = v.removeLast()                
23             }
24             if !v.isEmpty
25             {
26                 v.last!!.right = cur
27             }
28             v.append(cur)
29         }
30         return v.first!
31     }
32 }

160ms

 1 class Solution {
 2     func constructMaximumBinaryTree(_ nums: [Int]) -> TreeNode? {
 3         guard !nums.isEmpty else {
 4             return nil
 5         }
 6         var root = -1
 7         var rootIndex = 0
 8         var leftIndex = 0
 9         let rightIndex = nums.count - 1
10         while leftIndex <= rightIndex {
11             let left = nums[leftIndex]
12             if left > root {
13                 root = left
14                 rootIndex = leftIndex
15             }
16             leftIndex += 1
17         }
18         let tree = TreeNode(root)
19         tree.left = constructMaximumBinaryTree(Array(nums[0..<rootIndex]))
20         tree.right = constructMaximumBinaryTree(Array(nums[(rootIndex + 1)...]))
21         return tree
22    }
23 }

[Swift]LeetCode654. 最大二叉樹 | Maximum Binary Tree