Insert Node in a Binary Search Tree

Given a binary search tree and a new tree node, insert the node into 
the tree. You should keep the tree still be a valid binary search 
tree. 
Can you do it without recursion?

递归

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param root: The root of the binary search tree.
     * @param node: insert this node into the binary search tree
     * @return: The root of the new binary search tree.
     */
    TreeNode* insertNode(TreeNode* root, TreeNode* node) {
        // write your code here

        if(node==NULL){
            return root;
        }

        if(root==NULL){
            return node;
        }

        if(node->val<root->val){
            root->left=insertNode(root->left,node);
        }else{
            root->right=insertNode(root->right,node);
        }

        return root;
    }
};
  • 迭代
/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param root: The root of the binary search tree.
     * @param node: insert this node into the binary search tree
     * @return: The root of the new binary search tree.
     */
    TreeNode* insertNode(TreeNode* root, TreeNode* node) {
        // write your code here

        if(node==NULL){
            return root;
        }

        if(root==NULL){
            return node;
        }

        TreeNode *tempNode=root;
        while(tempNode){
            if(node->val<tempNode->val){
                if(tempNode->left==NULL){
                    tempNode->left=node;
                    return root;
                }
                tempNode=tempNode->left;
            }else{
                if(tempNode->right==NULL){
                    tempNode->right=node;
                    return root;
                }
                tempNode=tempNode->right;
            }
        }

        return root;
    }
};

### 如何使用二叉搜索树(BST)实现 A+B 操作 在 C 编程语言中,可以通过构建两个二叉搜索树(BST),分别表示集合 A 和 B 的元素,然后通过遍历其中一个 BST 并将其节点插入到另一个 BST 中来完成 A+B 操作。以下是详细的实现方法: #### 数据结构定义 首先需要定义一个简单的二叉搜索树节点的数据结构。 ```c typedef struct TreeNode { int value; struct TreeNode* left; struct TreeNode* right; } TreeNode; ``` #### 插入函数 为了向 BST 添加新元素,可以编写如下 `insert` 函数。 ```c TreeNode* createNode(int value) { TreeNode* newNode = (TreeNode*)malloc(sizeof(TreeNode)); newNode->value = value; newNode->left = NULL; newNode->right = NULL; return newNode; } void insert(TreeNode** root, int value) { if (*root == NULL) { *root = createNode(value); } else { if (value < (*root)->value) { insert(&((*root)->left), value); // Insert into the left subtree. } else if (value > (*root)->value) { insert(&((*root)->right), value); // Insert into the right subtree. } // If value == (*root)->value, do nothing since duplicates are not allowed in a set. } } ``` #### 合并操作 要执行 A+B 操作,即合并两棵 BST,可以从一棵树中提取所有元素并将它们逐个插入另一棵树中。 ```c // In-order traversal to extract elements from one tree and add them to another. void mergeTrees(TreeNode* sourceRoot, TreeNode** targetRoot) { if (sourceRoot != NULL) { mergeTrees(sourceRoot->left, targetRoot); // Traverse left subtree first. insert(targetRoot, sourceRoot->value); // Add current node's value to target tree. mergeTrees(sourceRoot->right, targetRoot); // Then traverse right subtree. } } ``` #### 主程序逻辑 假设我们已经初始化了两棵 BST 表示集合 A 和 B,则可以通过调用上述函数完成 A+B 操作。 ```c int main() { TreeNode* treeA = NULL; TreeNode* treeB = NULL; // Example: Adding values to Tree A. int arrayA[] = {5, 3, 7, 2, 4}; for (size_t i = 0; i < sizeof(arrayA)/sizeof(arrayA[0]); ++i) { insert(&treeA, arrayA[i]); } // Example: Adding values to Tree B. int arrayB[] = {6, 8, 1}; for (size_t i = 0; i < sizeof(arrayB)/sizeof(arrayB[0]); ++i) { insert(&treeB, arrayB[i]); } // Perform A + B by merging all nodes of treeB into treeA. mergeTrees(treeB, &treeA); // Now treeA contains all unique elements from both sets. return 0; } ``` 此代码片段展示了如何利用二叉搜索树的性质高效地进行集合并集运算[^1]。
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