LeetCode从上到下打印二叉树C++（1-3题）

题目链接：

「剑指 Offer」 - 学习计划 - 力扣（LeetCode）全球极客挚爱的技术成长平台

题目一：

/**
 * 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> levelOrder(TreeNode* root) {
        vector<int> ans;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()){
            int size = q.size();
            vector<int> level;
            while(size--){
                TreeNode* cur = q.front();
                q.pop();
                if(!cur){
                    continue;
                }
                    
                ans.push_back(cur->val);
                q.push(cur->left);
                q.push(cur->right);
                
            }

        }
    return ans;
    }
};

题目二：

模板笔记转自：

https://leetcode-cn.com/problems/binary-tree-level-order-traversal/solution/tao-mo-ban-bfs-he-dfs-du-ke-yi-jie-jue-by-fuxuemin/

BFS

/**
 * 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<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> ans;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()){
            int size = q.size();
            vector<int> level;
            while(size--){
                TreeNode* cur = q.front();
                q.pop();
                if(!cur){
                    continue;
                }
                    
                level.push_back(cur->val);
                q.push(cur->left);
                q.push(cur->right);
                
            }
            if(!level.empty())
            ans.push_back(level);
        }
    return ans;
    }
};

DFS

/**
 * 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<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> ans;
        
        dfs(ans,0,root);
        return ans;

        

    }
    void dfs(vector<vector<int>> &ans,int level,TreeNode* root){
            if(root == NULL) return;
            if(ans.size() == level - 1 || ans.size() == level){ 
                ans.push_back(vector<int> ());
            }
            ans[level].push_back(root->val);
            dfs(ans,level + 1 , root->left);
            dfs(ans,level + 1 , root->right);

        }
};

补充：C++中vector＜vector＜int＞ ＞

转自：关于C++中vector＜vector＜int＞ ＞的使用_xiziwen_code-CSDN博客

vector<vector<int> > A;
for(int i = 0; i < 2; ++i)  A.push_back(vector<int>());   
A[0].push_back(0);
A[0].push_back(1);
A[0].push_back(2);
A[1].push_back(3);
A[1].push_back(4);

//vector<vector<int> >A中的vector元素的个数
len = A.size();
//vector<vector<int> >A中第i个vector元素的长度
len = A[i].size();

//根据前面的插入，可知输出1。
printf("%d\n", A[0][1]);

题目三：没写新的，就是在题目2的bfs解法最后加了一个reverse

/**
 * 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<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> ans;
        queue<TreeNode*> q;
        q.push(root);
        int dep =0;
        while(!q.empty()){
            int size = q.size();
            vector<int> level;
            while(size--){
                TreeNode* cur = q.front();
                q.pop();
                if(!cur){
                    continue;
                }
                    
                level.push_back(cur->val);
                q.push(cur->left);
                q.push(cur->right);
                
            }
            if(!level.empty())
            ans.push_back(level);
        }
    for(int i = 1; i < ans.size(); i += 2)//ans.size()就是一共就有几行
            reverse(ans[i].begin(), ans[i].end());

    return ans;

    }
};