力扣hot100 二叉树的右视图 DFS BFS 层序遍历 递归

Problem: 199. 二叉树的右视图

思路

👩‍🏫 甜姨

💖 BFS

⏰ 时间复杂度: $O(n)$
🌎 空间复杂度: $O(n)$

class Solution {
   
    public List<Integer> rightSideView(TreeNode root) {
   
        List<Integer> res = new ArrayList<>();
        if (root == null) {
   
            return res;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
   
            int size = queue.size();
            for (int i = 0; i < size; i++) {
   
                TreeNode node = queue.poll();
                if (node.left != null) {
   
                    queue.offer(node.left);
                }
                if (node.right != null) {
   
                    queue.offer(node.right);
                }
                if (i == size - 1) {
     //将当前层的最后一个节点放入结果列表
                    res.add(node.val);
                }
            }
        }
        return res;
    }
}

💖 DFS

⏰ 时间复杂度: $O(n)$
🌎 空间复杂度: $O(n)$

/**
 * 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 {
	final List<Integer> ans = new ArrayList<>();

	public List<Integer> rightSideView(TreeNode root)
	{
		dfs(root, 0);
		return ans;
	}

	private void dfs(TreeNode root, int d)
	{
		if (root == null)
			return;
		if (d == ans.size())
			ans.add(root.val);
		dfs(root.right, d + 1);
		dfs(root.left, d + 1);
	}
}