DAY13 二叉樹part01

xzdmzrc發表於2024-07-29
二叉樹

今日任務

二叉樹的遞迴遍歷（前中後）

二叉樹的迭代遍歷（前中後）

二叉樹的統一迭代遍歷

二叉樹的層序遍歷（共十道題目）

完成情況

遞迴已掌握，程式碼略

迭代前中手寫一編，後和統一未學習

層序遍歷題目如下

102.二叉樹的層序遍歷

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 8  *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 9  *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
10  * };
11  */
12 class Solution {
13 public:
14     vector<vector<int>> levelOrder(TreeNode* root) {
15         queue<TreeNode*>que;
16         vector<vector<int>> result;
17         if(root!=NULL) que.push(root);
18         while(!que.empty())
19         {
20             int size=que.size();
21             vector<int> res;//寫到for迴圈外邊
22             for(int i=0;i<size;i++)
23             {
24                 
25                 TreeNode* cur=que.front();
26                 que.pop();
27                 res.push_back(cur->val);
28                 if(cur->left) que.push(cur->left);
29                 if(cur->right) que.push(cur->right);
30 
31             }
32             result.push_back(res);
33         }
　　　　　　　　reverse(result.begin(),result.end());//107.層序遍歷II只要加上反轉
34         return result;
35     }
36 };

199.二叉樹的右檢視

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 8  *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 9  *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
10  * };
11  */
12 class Solution {
13 public:
14     vector<int> rightSideView(TreeNode* root) {
15         queue<TreeNode*> que;
16         if (root != NULL) que.push(root);
17         vector<int> result;
18         while (!que.empty()) {
19             int size = que.size();
20             for (int i = 0; i < size; i++) {
21                 TreeNode* node = que.front();
22                 que.pop();
23                 if (i == (size - 1)) result.push_back(node->val); // 將每一層的最後元素放入result陣列中
24                 if (node->left) que.push(node->left);
25                 if (node->right) que.push(node->right);
26             }
27         }
28         return result;
29     
30     }
31 };

637.二叉樹的層平均值

注意精度要求

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 8  *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 9  *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
10  * };
11  */
12 class Solution {
13 public:
14     vector<double> averageOfLevels(TreeNode* root) {
15         queue<TreeNode*> que;
16         if (root != NULL) que.push(root);
17         vector<double> result;
18         while (!que.empty()) {
19             int size = que.size();
20             vector<double> vec;
21             for (int i = 0; i < size; i++) {
22                 TreeNode* node = que.front();
23                 que.pop();
24                 vec.push_back(node->val);
25                 if (node->left) que.push(node->left);
26                 if (node->right) que.push(node->right);
27             }
28             double sum=0;
29             for(int i=0;i<vec.size();i++)
30                 sum+=vec[i];
31             result.push_back(sum/vec.size());
32         }
33         return result;
34     }
35 };

429.N叉樹的層序遍歷

 1 /*
 2 // Definition for a Node.
 3 class Node {
 4 public:
 5     int val;
 6     vector<Node*> children;
 7 
 8     Node() {}
 9 
10     Node(int _val) {
11         val = _val;
12     }
13 
14     Node(int _val, vector<Node*> _children) {
15         val = _val;
16         children = _children;
17     }
18 };
19 */
20 
21 class Solution {
22 public:
23     vector<vector<int>> levelOrder(Node* root) {
24         queue<Node*>que;
25          vector<vector<int>> result;
26          if(root!=NULL) que.push(root);
27          while(!que.empty())
28          {
29              int size=que.size();
30              vector<int> res;//寫到for迴圈外邊
31              for(int i=0;i<size;i++)
32              {
33              
34                 Node* cur=que.front();
35                  que.pop();
36                  res.push_back(cur->val);
37                  if(!cur->children.empty())
38                  {
39                     vector<Node*> tmp=cur->children;
40                     for(int i=0;i<tmp.size();i++)
41                     {
42                         que.push(tmp[i]);
43                     }
44                  }
45             }
46             result.push_back(res);
47        }
48         return result;     
49     }
50 };

111.二叉樹的最小深度

 1 class Solution {
 2 public:
 3     int minDepth(TreeNode* root) {
 4         if (root == NULL) return 0;
 5         int depth = 0;
 6         queue<TreeNode*> que;
 7         que.push(root);
 8         while(!que.empty()) {
 9             int size = que.size();
10             depth++; // 記錄最小深度
11             for (int i = 0; i < size; i++) {
12                 TreeNode* node = que.front();
13                 que.pop();
14                 if (node->left) que.push(node->left);
15                 if (node->right) que.push(node->right);
16                 if (!node->left && !node->right) { // 當左右孩子都為空的時候，說明是最低點的一層了，退出
17                     return depth;
18                 }
19             }
20         }
21         return depth;
22     }
23 };

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