[LeetCode/Easy] 145. Binary Tree Postorder Traversal

1. Input

1) TreeNode root

 

2. Output

1) root를 후위순회한 결과를 리스트에 담아 반환

 

3. Constraint

1) root의 노드 수는 [0, 100]이다.

2) -100 <= Node.val <= 100

 

4. Example

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

 

5. Code

1) 첫 코드(2023/05/12)

/**
 * 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 List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> post = new ArrayList<Integer>();
        if(root!=null){
            post.addAll(postorderTraversal(root.left));
            post.addAll(postorderTraversal(root.right));
            post.add(root.val);
        }
        return post;
    }
}