DataStructure_Algorithm_HandBook_PreForLeetCode

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.

soluiton

这道题目也是一个构建的问题。 思想也是递归的形式去找 就是每次找到最大值的位置，为分割点，左右来建立数组

var constructMaximumBinaryTree = function(nums) {
  return dfs(0,nums.length-1,nums);
};
function dfs( l,  r, nums){
    if(l > r) return null;
    let max_index = get_MAX_index(nums,l,r);
    let node = new TreeNode(nums[max_index]); 
    node.left = dfs(l,max_index-1,nums);
    node.right = dfs(max_index+1,r,nums);
    return node; 
}

//找到 l 和 r 区间的最大值，返回下标
function get_MAX_index(nums,l,r){
    let max_index= -1; 
    let max_int = -1;
    for(let i = l; i<=r;i++){
        if(nums[i]>max_int){
            max_int = nums[i];
            max_index = i; 
        }
    }
    return max_index; 
}