剑指 Offer 32 从上到下打印二叉树IIIIII

Posted 2022-05-05 Billy Miracle

I

菜鸡思路：

数组的层次遍历，放到数组里即可。

菜鸡代码：

/**
 * 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> result;
        queue<TreeNode*> que;
        if(root != NULL) 
            que.push(root);
        
        while(!que.empty()) 
            int size = que.size();
            for(int i = 0; i < size; i++) 
                TreeNode* node = que.front();
                que.pop();
                result.push_back(node->val);
                if(node->left != NULL) 
                    que.push(node->left);
                
                if(node->right != NULL) 
                    que.push(node->right);
                
            
        
        return result;
    
;

II

菜鸡思路：

虽然还是层次遍历，但是要求分层返回，有些棘手，那么可以这样，每次判断一下一层有几个，然后针对性返回该层。

菜鸡代码：

/**
 * 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>> result;
        queue<TreeNode *> que;
        if (root) 
            que.push(root);
        
        while (que.size()) 
            vector<int> row;
            int size = que.size();
            //记录下这一层有几个
            for (int i = 0; i < size; ++i) 
            //针对每层的节点数分层操作
                TreeNode *node = que.front();
                que.pop();
                row.push_back(node->val);
                if (node->left) 
                    que.push(node->left);
                
                if (node->right) 
                    que.push(node->right);
                
            
            result.push_back(row);
        
        return result;
    
;

III

菜鸡思路：

虽然还是层次遍历、分层返回，但是要求的是要针对性的对一些层倒序，直接在上一题代码改造，判断一下是第几层，偶数层则数组倒序。只比上一题多了一个if条件块。

菜鸡代码：

/**
 * 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>> result;
        queue<TreeNode *> que;
        if (root) 
            que.push(root);
        
        while (que.size()) 
            vector<int> row;
            int size = que.size();
            for (int i = 0; i < size; ++i) 
                TreeNode *node = que.front();
                que.pop();
                row.push_back(node->val);
                if (node->left) 
                    que.push(node->left);
                
                if (node->right) 
                    que.push(node->right);
                
            
            if (result.size() % 2) 
                reverse(row.begin(), row.end());
            
            result.push_back(row);
        
        return result;
    
;

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