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].

==============

利用BFS遍历二叉树的方法,

queue<TreeNode*> q;

curr/next分别记录当前层要遍历的节点数量,下层要遍历的节点数量

====

/**
 * 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> re;
        if(root==nullptr) return re;
        help_rightSv(root,re);

        for(auto i:re){
            cout<<i<<" ";
        }cout<<endl;
        return re;
    }

    void help_rightSv(TreeNode *root,vector<int> &path){
        queue<TreeNode*> q;
        int curr = 1;
        int next = 0;
        int level = 0;
        q.push(root);
        while(!q.empty()){
            if(curr>0){
                TreeNode *tmp = q.front();
                if(path.size()==level) {
                    path.push_back(tmp->val);
                }
                q.pop();
                curr--;
                if(tmp->right!=nullptr){
                    q.push(tmp->right);
                    next++;
                }
                if(tmp->left!=nullptr){
                    q.push(tmp->left);
                    next++;
                }
            }else{
                curr = next;
                next = 0;
                level++;
            }
        }///while
    }
};