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701. Insert into a Binary Search Tree

MediumOpen on LeetCodeProblem statement

Problem Statement

701. Insert into a Binary Search Tree

Medium

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:

C++

Source file
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* insertIntoBST(TreeNode* root, int val) {
        //Exception Handling in case of Empty BST
        if (!(root))
            return new TreeNode(val);
        
        bool childDir; //Direction of Child - Left(false) or Right(true)
        // Traverse given BST
        auto ptr = root;
        auto prev = ptr;
        while(ptr){
            prev = ptr;
            if (val>ptr->val){
                ptr = ptr->right;
                childDir = true;
            }
            else{ 
                ptr=ptr->left;
                childDir = false;
            }
        }
        // Insert `val` in correct child position
        if(childDir)
            prev->right = new TreeNode(val);
        else prev->left = new TreeNode(val);
        return root;
    }
};