4.4-按行输出二叉树（same in LeetCode）

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Given a binary search tree, design an algorithm which creates a linked list of all the nodes at each depth (eg, if you have a tree with depth D, you’ll have D linked lists).

LeetCode:https://oj.leetcode.com/submissions/detail/3391867/

#include <iostream>#include <vector>#include <queue>using namespace std;struct TreeNode{    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};vector<vector<int> > levelOrder_2vec(TreeNode *root){    vector<vector<int> > ans;    vector<int> level;    vector<TreeNode *> cur,next;    cur.push_back(root);    while(!cur.empty())    {        level.clear();        next.clear();        for(int i=0;i<cur.size();i++)        {            level.push_back(cur[i]->val);            if(cur[i]->left)              next.push_back(cur[i]->left);            if(cur[i]->right)              next.push_back(cur[i]->right);        }        ans.push_back(level);        cur=next;    }    return ans;}vector<vector<int> > levelOrder_queue(TreeNode *root){    vector<vector<int> > ans;    vector<int> level;    queue<TreeNode *> myqueue;    myqueue.push(root);    int cur_count=1;    while(!myqueue.empty())    {        level.clear();        int next_count=0;        for(int i=0;i<cur_count;i++)        {            TreeNode * tmp = myqueue.front();            level.push_back(tmp->val);            myqueue.pop();            if(tmp->left)             {                 myqueue.push(tmp->left);                 next_count++;             }            if(tmp->right)             {                 myqueue.push(tmp->right);                 next_count++;             }        }        ans.push_back(level);        cur_count = next_count;    }    return ans;}int main(){    TreeNode * a = new TreeNode(1);    TreeNode * b = new TreeNode(2);    TreeNode * c = new TreeNode(3);    TreeNode * d = new TreeNode(4);    TreeNode * e = new TreeNode(5);    TreeNode * f = new TreeNode(6);    a->left=b;    a->right=c;    b->right=d;    c->left=e;    c->right=f;    vector<vector<int> > ans = levelOrder_2vec(a);    for(int i=0; i<ans.size(); i++)    {        for(int j=0; j<ans[i].size(); j++)            cout << ans[i][j];        cout << endl;    }    cout <<"---------------------" <<endl;    vector<vector<int> > ans2 = levelOrder_queue(a);    for(int i=0; i<ans2.size(); i++)    {        for(int j=0; j<ans2[i].size(); j++)            cout << ans2[i][j];        cout << endl;    }    return 0;}

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