代码随想录算法训练营第十七天| 654.最大二叉树 617.合并二叉树 700.二叉搜索树中的搜索 98.验证二叉搜索树

 654.最大二叉树 

比较简单，直接上代码

 1 TreeNode* constructMax_cursor(vector<int>& nums)
 2 {
 3     if (nums.size() == 0) return NULL;
 4     //getMaxNum
 5     int index = 0;
 6     int max_ = INT_MIN;
 7     for (int i = 0; i < nums.size(); i++)
 8     {
 9         if (max_ < nums[i])
10         {
11             index = i;
12             max_ = nums[i];
13         }
14     }
15 
16     TreeNode* curNode = new TreeNode(max_);
17 
18     //分割数组
19     vector<int> left = vector<int>(nums.begin(), nums.begin() + index);
20 
21     vector<int> right = index + 1 != nums.size() ?
22         vector<int>(nums.begin() + index + 1, nums.end()) : vector<int>(nums.end(), nums.end());
23 
24     
25     curNode->left = constructMax_cursor(left);
26     curNode->right = constructMax_cursor(right);
27 
28     return curNode;
29 }
30 TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
31     if (nums.size() == 0) return NULL;
32 
33     return constructMax_cursor(nums);
34 }

 617.合并二叉树

代码

 1 TreeNode* mergeTrees_cur(TreeNode* root1, TreeNode* root2)
 2 {
 3     TreeNode* node = new TreeNode();
 4     if (!root1 && !root2) return NULL;
 5     else if (root1 && root2)
 6     {
 7         node->val = root1->val + root2->val;
 8         node->left = mergeTrees_cur(root1->left, root2->left);
 9         node->right = mergeTrees_cur(root1->right, root2->right);
10     }
11     else if (root1 && !root2)
12     {
13         node->val = root1->val;
14         node->left = mergeTrees_cur(root1->left, NULL);
15         node->right = mergeTrees_cur(root1->right, NULL);
16     }
17     else if (!root1 && root2)
18     {
19         node->val = root2->val;
20         node->left = mergeTrees_cur(NULL, root2->left);
21         node->right = mergeTrees_cur(NULL, root2->right);
22     }
23 
24     return node;
25 }
26 TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) 
27 {
28     if (!root1 && !root2) return NULL;
29     return mergeTrees_cur(root1, root2);
30 }

 700.二叉搜索树中的搜索 

思路：

因为二叉搜索数是一个有序树，所以可以直接根据和根节点的大小，取匹配

代码：

 1 TreeNode* searchBST(TreeNode* root, int val) {
 2     TreeNode* result = NULL;
 3     if (!root) return result;
 4 
 5     while (root != NULL)
 6     {
 7         if (root->val == val)
 8         {
 9             return root;
10         }
11         else if (root->val > val)
12         {
13             root = root->left;
14         }
15         else
16         {
17             root = root->right;
18         }
19     }
20     return result;
21 }

 98.验证二叉搜索树 

中序遍历，保证从小到大

代码：

 1 bool isValidBST(TreeNode* root) {
 2     if (!root) return false;
 3     vector<int> vals;
 4     stack<TreeNode*> selected;
 5     selected.push(root);
 6     while (!selected.empty())
 7     {
 8         auto cur_ = selected.top();
 9         if (cur_->left)
10         {
11             selected.push(cur_->left);
12             cur_->left = NULL;
13         }
14         else
15         {
16             if (vals.size()>0&&cur_->val <= vals.back())
17             {
18                 return false;
19             }
20             vals.push_back(cur_->val);
21             selected.pop();
22             if (cur_->right) selected.push(cur_->right);
23         }
24     }
25 
26     return true;
27 }

 

 

posted @ 2023-06-26 11:31  博二爷  阅读(4)  评论(0)  编辑  收藏  举报