[LeetCode/Easy] 530. Minimum Absolute Difference in BST

1. Input

1) TreeNode root

 

2. Output

1) root에 있는 두 노드의 값의 차가 가장 작은 것을 반환

- 차는 절댓값으로 계산한다.

 

3. Constraint

1) 노드의 수는 [2, 10^4]이다.

2) 0 <= Node.val <= 10^5

 

4. Example

Input: root = [4,2,6,1,3] -> Output: 1

 

5. Code

1) 첫 코드(2023/05/23)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int getMinimumDifference(TreeNode root) {
        int min = Integer.MAX_VALUE;
        List<Integer> list = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while(!stack.empty()){
            TreeNode cur = stack.pop();
            list.add(cur.val);
            if(cur.left!=null)
                stack.push(cur.left);
            if(cur.right!=null)
                stack.push(cur.right);
        }
        list.sort(Comparator.naturalOrder());
        for(int i=0 ; i<list.size()-1 ; i++)
            min = Math.min(min, list.get(i+1)-list.get(i));
        return min;
    }
}