LeetCode 199. Binary Tree Right Side View

原题链接在这里：https://leetcode.com/problems/binary-tree-right-side-view/

题目：

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]. 

题解：

与Binary Tree Level Order Traversal相似，通过BFS一层一层扫树，这里是仅添加最右面的一个点，就是curCount = 1 的时候.

Note: 当curCount == 0 时, curCount = nextCount, 不要忘记把nextCount归0.

Time Complexity: O(n). n是tree的node总数.

Space: O(n), que的大小.

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root == null){
            return res;
        }
        
        LinkedList<TreeNode> que = new LinkedList<TreeNode>();
        que.add(root);
        int curCount = 1;
        int nextCount = 0;
        while(!que.isEmpty()){
            TreeNode cur = que.poll();
            curCount--;
            
            if(cur.left != null){
                que.add(cur.left);
                nextCount++;
            }
            if(cur.right != null){
                que.add(cur.right);
                nextCount++;
            }
            
            if(curCount == 0){
                res.add(cur.val);
                curCount = nextCount;
                nextCount = 0;
            }
        }
        
        return res;
    }
}