Leetcode 654. Maximum Binary Tree

You are given an integer array nums with no duplicates. A maximum binary tree can be built recursively from nums using the following algorithm:

1. Create a root node whose value is the maximum value in nums.
2. Recursively build the left subtree on the subarray prefix to the left of the maximum value.
3. Recursively build the right subtree on the subarray suffix to the right of the maximum value.

Return the maximum binary tree built from nums.

Example 1:

Input: nums = [3,2,1,6,0,5]
Output: [6,3,5,null,2,0,null,null,1]
Explanation: The recursive calls are as follow:
- The largest value in [3,2,1,6,0,5] is 6. Left prefix is [3,2,1] and right suffix is [0,5].
    - The largest value in [3,2,1] is 3. Left prefix is [] and right suffix is [2,1].
        - Empty array, so no child.
        - The largest value in [2,1] is 2. Left prefix is [] and right suffix is [1].
            - Empty array, so no child.
            - Only one element, so child is a node with value 1.
    - The largest value in [0,5] is 5. Left prefix is [0] and right suffix is [].
        - Only one element, so child is a node with value 0.
        - Empty array, so no child.

Example 2:

Input: nums = [3,2,1]
Output: [3,null,2,null,1]

Constraints:

1 <= nums.length <= 1000
0 <= nums[i] <= 1000
All integers in nums are unique.

解析：给定一个不重复元素的数组，构建二叉树。

1.根节点是数组中数值最大的元素。

2.左子树数值来源于数组中最大值左边的数字

3.右子树数值来源于数组中最大值右边的数字

每个二叉树节点都可以认为是一棵子树的根节点，对于根节点，首先要做的当然是把想办法把自己先构造出来，然后想办法构造自己的左右子树。

所以，我们要遍历数组把找到最大值maxVal，从而把根节点 root做出来，然后对maxVal左边的数组和右边的数组进行递归构建，作为root的左右子树。

代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
        return build(nums, 0, nums.size()-1);
    }
    TreeNode* build(vector<int>& nums, int low, int high){
        if(low > high){
            return NULL;
        }         
        int index=-1, maxValue=INT_MIN;
        for(int i = low; i <= high; i++){
            if(maxValue < nums[i]){
                index = i;
                maxValue = nums[i];
            }
        }
        TreeNode* root = new TreeNode(maxValue);
        root->left = build(nums, low, index-1);
        root->right = build(nums, index+1, high);
        return root;
    }
};