199. 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].
BFS的方法，即层序遍历
代码如下：
注意front和size的事先存留

/**
 * 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>vec;
        if(root==NULL)return vec;
        queue<TreeNode*>nQueue;
        TreeNode *node=root;
        nQueue.push(node);
        while(!nQueue.empty()){
            int size=nQueue.size();//这里注意size要保存下来，因为不然后面nQueue大小会变的啊。
            for(int i=0;i<size;i++){
                node=nQueue.front();//queue用front和back
                nQueue.pop();
                if(i==0){
                    vec.push_back(node->val);
                }
                if(node->right){
                    nQueue.push(node->right);
                }
                if(node->left){
                     nQueue.push(node->left);
                }
            }
        }
        return vec;
    }
};