[LeetCode] Binary Tree Right Side View

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:

分析：层次遍历，取最右面的元素：

方法一：双q法：

 

class Solution {
    public:
        vector<int>  rightSideView(TreeNode *root)
        {
            queue<TreeNode*> q1;
            queue<TreeNode*> q2;
            vector<int>  res;

            if(root != NULL)
            {
                q1.push(root);
            }

            while(!q1.empty())
            {
                TreeNode * p = q1.front();
                q1.pop();

                if(p->left)
                    q2.push(p->left);
                if(p->right)
                    q2.push(p->right);

                if(q1.empty() /*&& !q2.empty()*/)
                {
                    res.push_back(p->val);
                    swap(q1, q2);
                }
            }
            return res;
        }
};

 

方法二：单q+ NULL标记level结束法

/**
 * Definition for a binary tree node.
 * 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> rtn;
            if(root == NULL)
                return rtn;

            queue<TreeNode*> q;

            TreeNode* pre = NULL;

            q.push(root);
            q.push(NULL);//mark the end of this level

            while(!q.empty())
            {   
                TreeNode * p = q.front();
                q.pop();

                if(p == NULL) 
                {   
                    rtn.push_back(pre->val);

                    if(!q.empty())// some elements still exist in q.
                        q.push(NULL);//mark end of this level

                }   
                else
                {   
                    if(p->left)
                        q.push(p->left);
                    if(p->right)
                        q.push(p->right);
                }   
                pre = p;
            }

            return rtn;
        }
};