[Leetcode 701] Insert into a Binary Search Tree

Description:

You are given the root node of a binary search tree (BST) and a value to insert into the tree. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.

Notice that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return any of them.

Example 1:

Input: root = [4,2,7,1,3], val = 5
Output: [4,2,7,1,3,5]
Explanation: Another accepted tree is:

Example 2:

Input: root = [40,20,60,10,30,50,70], val = 25
Output: [40,20,60,10,30,50,70,null,null,25]

Example 3:

Input: root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
Output: [4,2,7,1,3,5]

Constraints:

• The number of nodes in the tree will be in the range [0, 104].
• -108 <= Node.val <= 108
• All the values Node.val are unique.
• -108 <= val <= 108
• It’s guaranteed that val does not exist in the original BST.

C code solution (recursive):

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* insertIntoBST(struct TreeNode* root, int val){
    
    struct TreeNode* newNode = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    newNode->val = val;
    newNode->left = NULL;
    newNode->right = NULL;
    
    if(!root){
        return newNode;
    }
    
    if(val < root->val)
        root->left = insertIntoBST(root->left, val);
    else if(val > root->val)
        root->right = insertIntoBST(root->right, val);
    
    return root;
}

C code solution (iterative):

struct TreeNode* insertIntoBST(struct TreeNode* root, int val){
    
    struct TreeNode* newNode = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    newNode->val = val;
    newNode->left = NULL;
    newNode->right = NULL;
    
    if(!root){
        return newNode;
    }
    
    
    struct TreeNode* prev;
    struct TreeNode* curr = root;
    int is_left = 0;
    //trace to the bottom node
    while(curr){
        if(val < curr->val){
            prev = curr;
            curr = curr->left;
            is_left = 1;
        }
        else{
            prev = curr;
            curr = curr->right;
            is_left = 0;
        }
    }
    //decide the node should be at left or right of the bottom node
    if(is_left)
        prev->left = newNode;
    else
        prev->right = newNode;
    
    return root;
}

A very clean code solution:

struct TreeNode* insertIntoBST(struct TreeNode* root, int val){
    
    struct TreeNode* newNode = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    newNode->val = val;
    newNode->left = NULL;
    newNode->right = NULL;
    
    if(!root){
        return newNode;
    }
    
    struct TreeNode** pp = &root;
    
    while (*pp != NULL) {
        if (val > (*pp)->val)
            pp = &(*pp)->right;
        else
            pp = &(*pp)->left;
    }
    
    *pp = newNode;
    
    return root;
}