701. Insert into a Binary Search Tree

Problem 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.

Examples:

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

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

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

Python Solution

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
        if not root:
            return TreeNode(val)
            
        if val < root.val:
            root.left = self.insertIntoBST(root.left, val)
        else:
            root.right = self.insertIntoBST(root.right, val)
            
        return root

Implementation Notes:

• Uses recursive approach to find the correct insertion point
• Time complexity: O(H) where H is the height of the tree
• Space complexity: O(H) for recursion stack

Java Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode insertIntoBST(TreeNode root, int val) {
        if (root == null) {
            return new TreeNode(val);
        }
        
        if (val < root.val) {
            root.left = insertIntoBST(root.left, val);
        } else {
            root.right = insertIntoBST(root.right, val);
        }
        
        return root;
    }
}

C++ Solution

/**
 * 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) {
        if (!root) {
            return new TreeNode(val);
        }
        
        if (val < root->val) {
            root->left = insertIntoBST(root->left, val);
        } else {
            root->right = insertIntoBST(root->right, val);
        }
        
        return root;
    }
};

JavaScript Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} val
 * @return {TreeNode}
 */
var insertIntoBST = function(root, val) {
    if (!root) {
        return new TreeNode(val);
    }
    
    if (val < root.val) {
        root.left = insertIntoBST(root.left, val);
    } else {
        root.right = insertIntoBST(root.right, val);
    }
    
    return root;
};

C# Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public TreeNode InsertIntoBST(TreeNode root, int val) {
        if (root == null) {
            return new TreeNode(val);
        }
        
        if (val < root.val) {
            root.left = InsertIntoBST(root.left, val);
        } else {
            root.right = InsertIntoBST(root.right, val);
        }
        
        return root;
    }
}

Implementation Notes:

• Uses recursive approach to find the correct insertion point
• Takes advantage of BST property to decide left or right traversal
• Time complexity: O(H) where H is the height of the tree
• Space complexity: O(H) for recursion stack