199. Binary Tree Right Side View

DFS

比较每个节点的深度及用字符串纪录它的位置；

也可以用用层次遍历，并且记下每层的节点；

struct Node{    int val;    string rt;    Node(int x,string y):val(x),rt(y){}};class Solution {public:    void DFS(TreeNode* root,string rt,int depth,vector<Node> &vt)    {        if(!root)            return;        if(depth<vt.size())        {            if(rt>vt[depth].rt)            {                vt[depth].rt=rt;                vt[depth].val=root->val;            }        }        else        {            Node node(root->val,rt);            vt.push_back(node);        }        DFS(root->left,rt+"1",depth+1,vt);        DFS(root->right,rt+"2",depth+1,vt);    }        vector<int> rightSideView(TreeNode* root) {        vector<int> vec;        vector<Node> vt;        DFS(root, "", 0, vt);        for(int i=0;i<vt.size();i++)            vec.push_back(vt[i].val);        return vec;    }    };

BFS

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