654. Maximum Binary Tree 最大节点劈开,然后左边、右边排序

时间:2022-12-31 10:09:38

[抄题]:

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

[暴力解法]:

时间分析:

空间分析:

[优化后]:

时间分析:

空间分析:

[奇葩输出条件]:

[奇葩corner case]:

TreeNode返回空值对应的是null

[思维问题]:

不知道dc怎么写:recursion的条件是start end就行了。

[英文数据结构或算法,为什么不用别的数据结构或算法]:

[一句话思路]:

[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

[画图]:

[一刷]:

  1. helper函数中建了节点,很显然主函数中就不用再建了
  2. buildtree必须写root.left root.right

[二刷]:

[三刷]:

[四刷]:

[五刷]:

[五分钟肉眼debug的结果]:

[总结]:

buildtree必须写root.left root.right,recursion的条件是start end就行了。

[复杂度]:Time complexity: O(nlgn) Space complexity: O(n)

[算法思想:迭代/递归/分治/贪心]:

[关键模板化代码]:

[其他解法]:

[Follow Up]:

[LC给出的题目变变变]:

[代码风格] :

[是否头一次写此类driver funcion的代码] :

[潜台词] :

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode constructMaximumBinaryTree(int[] nums) {
//corner case
if (nums == null || nums.length == 0) return null; //utilize
return constructHelper(nums, 0, nums.length - 1);
} public TreeNode constructHelper(int[] nums, int start, int end) {
//exit requirement
if (start > end) return null; //find the max
int maxIdx = start;
for (int i = start + 1; i <= end; i++) {
if (nums[i] > nums[maxIdx])
maxIdx = i;
}
TreeNode root = new TreeNode(nums[maxIdx]); //recursive in left and right
root.left = constructHelper(nums, start, maxIdx - 1);
root.right = constructHelper(nums, maxIdx + 1, end); //return
return root;
}
}