[LeetCode]Binary Tree Right Side View

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

题意描述：

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：

这道题非常有趣，是找二叉树从右侧看的时候能看到的数，其实思路也是很简单，即返回每一层的最右边的一个数就好了，在二叉树的层序遍历的代码上稍作修改即可。对二叉树的层序遍历不太清楚的朋友，可看我之前的博客《二叉树完全总结》。具体没什么好说的，直接上代码：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root==null)
            return res;
        Queue<TreeNode> Q = new LinkedList<TreeNode>(); 
        Q.offer(root);
        int curLevel = 1;
        res.add(root.val);
        while(!Q.isEmpty()){
            int levelSize = Q.size();
            int count = 0;
            TreeNode last = null;
            while(count<levelSize){
                TreeNode p = Q.poll();
                if(p.left!=null){
                    last = p.left;
                    Q.offer(p.left);
                }
                if(p.right!=null){
                    last = p.right;
                    Q.offer(p.right);
                }
                count++;
            }
            if(last!=null)
                res.add(last.val);
            
        }
        return res;
    }
}