LeetCode - 654. Maximum Binary Tree

654. Maximum Binary Tree - medium

문제

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.

제한사항

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

입출력 예

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

      6
    /   \
   3     5
    \    / 
     2  0   
       \
        1

풀이

• Tree, Divide and Conquer

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:    
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
        if(nums.empty())
           return nullptr;
        
        // 현재 vector의 최대값 index를 검색
        // 찾은 index를 기반으로 최대값을 tree의 root으로 새로 node를 생성 
        auto maxValIndex = max_element(nums.begin(), nums.end()) - nums.begin();
        TreeNode* node = new TreeNode(nums[maxValIndex]);
        
        // 최대값 index의 왼쪽의 sub vector를 재귀를 통해
        // 하위 left child node를 찾아 tree를 구성
        vector<int> leftNums(nums.begin(), nums.begin() + maxValIndex);
        node->left = constructMaximumBinaryTree(leftNums);

        // 최대값 index의 오른쪽의 sub vector를 재귀를 통해
        // 하위 right child node를 찾아 tree를 구성
        vector<int> rightNums(nums.begin() + maxValIndex + 1, nums.end());
        node->right = constructMaximumBinaryTree(rightNums);
        
        return node;
    }
};