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方法一 BFS

使用二叉树层序遍历的思想 并用size来记录每一层的节点个数，当size==1时就是每一层的最后一个

/** * 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
   
     rightSideView(TreeNode root) {
           List
    
      list = new ArrayList
     
      ();        if(root == null) {
               return list;        }        Queue
      
        queue = new LinkedList<>();        queue.offer(root);                while(!queue.isEmpty()) {
               int size = queue.size();            //保存某一层的节点个数            while(size > 0) {
                   if(size == 1) {
                       //说明到某一层的最右边了                    list.add(queue.peek().val);                }                TreeNode cur = queue.poll();                size--;                if(cur.left != null) {
                       queue.offer(cur.left);                }                if(cur.right != null) {
                       queue.offer(cur.right);                }            }                    }        return list;    }}
      
     
    
   

方法二 DFS

/** * 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 {
       List
   
     list = new ArrayList<>();    public List
    
      rightSideView(TreeNode root) {
           if(root == null) {
               return list;        }        dfs(root, 0);        return list;    }    public void dfs(TreeNode root, int depth) {
           if(root == null) {
               return;        }        if(list.size() == depth) {
               list.add(root.val);        }        //每进入一次递归  深度加一        depth++;        dfs(root.right, depth);        dfs(root.left, depth);    }}
    
   

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