199. Binary Tree Right Side View java solutions

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

For example:
Given the following binary tree,

   1            <---
 /   
2     3         <---
      
  5     4       <---

You should return [1, 3, 4].

Credits:
Special thanks to @amrsaqr for adding this problem and creating all test cases.

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<Integer> rightSideView(TreeNode root) {
12         List<Integer> ans = new ArrayList<Integer>();
13         addright(ans,0,root);
14         return ans;
15     }
16     
17     public void addright(List<Integer> ans, int high, TreeNode root){
18         if(root == null) return;
19         if(ans.size() == high) ans.add(root.val);// 每一层加入一个元素
20         addright(ans,high+1,root.right);
21         addright(ans,high+1,root.left);
22     }
23 }

该题可以用递归解决，每次加入元素从最右边开始，每层加入一个。左子树能加入的情况就是high 高度大于右子树的情况。

如：

2 5

中元素4所处的情况。虽然代码上看是递归，实际上程序执行时间是O(n)复杂度。

解法2：

该题也可以使用按照层序遍历的方式，每层把最右节点的值加入list即可。

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<Integer> rightSideView(TreeNode root) {
12         List<Integer> ans = new ArrayList<Integer>();
13         if(root == null) return ans;
14         Deque<TreeNode> q = new LinkedList<TreeNode>();
15         int current = 1,next = 0;//使用current 和next来标记当前层后下一层的元素个数
16         TreeNode node;
17         q.addLast(root);
18         while(q.size() > 0){
19             node = q.removeFirst();
20             current--;
21             if(node.left != null){
22                 q.addLast(node.left);
23                 next++;
24             }
25             if(node.right != null){
26                 q.addLast(node.right);
27                 next++;
28             }
29             if(current == 0){
30                 ans.add(node.val);//每次加入该层最右元素
31                 current = next;
32                 next = 0;
33             }
34         }
35         return ans;
36     }
37     
38 }

实际运行时间：

不如解法一。