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].

一般是用bfs的做法，每次将队尾也就是每层最右边的元素push进res里面。

class Solution {public:    vector<int> rightSideView(TreeNode* root) {                vector<int> res;         queue<TreeNode*> tmp;        if(!root) return res;        tmp.push(root);        while(!tmp.empty()){            res.push_back(tmp.back()->val);            for(int i=tmp.size(); i!=0; --i){                TreeNode* t=tmp.front();                tmp.pop();                if(t->left) tmp.push(t->left);                if(t->right) tmp.push(t->right);            }                    }                return res;    }    

但是有种dfs递归的做法也很巧妙。省去了队列的空间。主要思想是每层只push进一个节点，且函数递归调用时先遍历右边的节点，因此当遍历到左边的节点时，如果存在右节点的话，res的size和curDepth是不相等的，因此也不会被push进res中。

class Solution {public:    vector<int> rightSideView(TreeNode* root) {                vector<int> res;        helper(res, root, 0);        return res;    }        void helper(vector<int>& res, TreeNode* cur, int curDepth){        if(!cur) return ;        if(res.size()==curDepth){            res.push_back(cur->val);        }                helper(res, cur->right, curDepth+1);        helper(res, cur->left, curDepth+1);    }};