654. Maximum Binary Tree

发表于 | 分类于 | |

654. Maximum Binary Tree

题目描述

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

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

1
2
3
4
5
6
7
8
9
10
11
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:
The size of the given array will be in the range [1,1000].

题目大意

通过数组建立最大二叉树

解题思路

1、找到数组最大值作为根
2、挑选左半数组内的最大值作为左子树的根
3、挑选右半数组内的最大值作为右子树的根
以此类推
ps:其实就是题目描述

代码

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
32
33
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
     func constructMaximumBinaryTree(_ nums: [Int]) -> TreeNode? {
        func construct(start: Int, end: Int) -> TreeNode? {
            guard start <= end else { return nil }
            var index = start
            var max = nums[start]
            for i in start...end {
                if nums[i] > max {
                    index = i
                    max = nums[i]
                }
            }
            let node = TreeNode(max)
            node.left = construct(start: start, end: index - 1)
            node.right = construct(start: index + 1, end: end)
            return node
        }
        return construct(start: 0, end: nums.count - 1)
    }
}
0%