10.二叉树的右视图

• 2.11

  • 秒了

• 1.29

  • 没秒，应该是 i= size-1

• 1.17

  • 依旧秒，但是有一些健壮性判断没做，如果手撕的话有 bug

• 11.11

  • 真变强了,知道是层序遍历就立马秒了(没细看答案)

/**
 * 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> rightSideView(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        if(root==null) return list;
        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(i == size-1){
                    list.add(node.val);
                }
                if(node.left!=null) queue.offer(node.left);
                if(node.right!=null) queue.offer(node.right);
            }
        }
        return list;
    }
}

• 就是在层序遍历中,检查会否为队列的最后一个size-1