Leetcode: 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]​​.

思路分析：

二叉树的层次遍历，在遍历过程中输出每一层最右边的节点。

我们可以在层次遍历过程中先访问右边节点，然后访问做变节点，访问结果入队列，然后每一层最前边的节点放入结果集合中。

C++参考代码：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution
{
public:
    vector<int> rightSideView(TreeNode *root)
    {
        vector<int> result;
        if (!root) return result;
        queue<TreeNode*> nodeQueue;
        TreeNode *node = root;
        nodeQueue.push(node);
        while (!nodeQueue.empty())
        {
            queue<TreeNode*>::size_type size = nodeQueue.size();
            //每次依次取出这一层的节点，然后将改元素的左右节点入队列（如果是第一个节点，则放入结果集合中）
            for (size_t i = 0; i < size; ++i)
            {
                node = nodeQueue.front();
                nodeQueue.pop();
                if (i == 0) result.push_back(node->val);
                //注意是先右节点后左节点
                if (node->right) nodeQueue.push(node->right);
                if (node->left) nodeQueue.push(node->left);
            }
        }
    
        return result;
    }
};

C#参考代码：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution
{
    public IList<int> RightSideView(TreeNode root)
    {
        IList<int> result = new List<int>();
        if (root == null) return result;
        Queue<TreeNode> nodeQueue = new Queue<TreeNode>();
        TreeNode node = root;
        nodeQueue.Enqueue(node);
        while (nodeQueue.Count != 0)
        {
            int size = nodeQueue.Count;
            for (int i = 0; i < size; ++i)
            {
                node = nodeQueue.Dequeue();
                if (i == 0) result.Add(node.val);
                if (node.right != null) nodeQueue.Enqueue(node.right);
                if (node.left != null) nodeQueue.Enqueue(node.left);
            }
        }
        return result;
    }
}