💡 LeetCode 101 - Symmetric Tree

문제

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

입출력 예제

✅ 예제 1

Input: root = [1,2,2,3,4,4,3]
Output: true

✅ 예제 2

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

제약조건

• The number of nodes in the tree is in the range [1, 1000]. 
• -100 <= Node.val <= 100

작성 코드

/**
 * 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 boolean isSymmetric(TreeNode root) {
		// 1. 빈 노드가 주어졌다면 참 처리
		if (root == null) return true;
		
		// 2. 반환
		return dfs(root.left, root.right);
	}
	
	/**
	 * DFS
	 */
	public boolean dfs(TreeNode leftChildNode, TreeNode rightChildNode) {
		// 1. 유효성 체크
		if (leftChildNode == null && rightChildNode == null) return true;
		if (leftChildNode == null || rightChildNode == null) return false;
		if (leftChildNode.val != rightChildNode.val) return false;
		
		// 2. DFS 처리
		return dfs(leftChildNode.left, rightChildNode.right) && dfs(leftChildNode.right, rightChildNode.left);
	}
}

회고

• BFS와 DFS 모두 적용 할 수 있는 경우 어떤 방법이 더욱 적합한 지도 체크해야겠다.