You are given therootnode of a binary search tree (BST) and avalueto insert into the tree. Returnthe root node of the BST after the insertion. It isguaranteedthat the new value does not exist in the original BST.

Noticethat there may existmultiple valid ways for theinsertion, as long as the tree remains a BST after insertion. You can returnany 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 inthe tree will be in the range[0,104].
• -108<= Node.val <= 108
• All the valuesNode.valareunique.
• -108<= val <= 108
• It'sguaranteedthatvaldoes not exist in the original BST.

M1: recursion, time = O(height), space = O(height) call stack

class Solution {
    public
 
 TreeNode insertIntoBST(TreeNode root, int val) {
        TreeNode node = new TreeNode(val);
        if(root == null) {
            return node;
        }
        if(root.val < val) {
            root.right = insertIntoBST(root.right, val);
        } else {
            root.left = insertIntoBST(root.left, val);
        }
        return root;
    }
}

M2: iterative, time = O(height), space = O(1)

class Solution {
    public TreeNode insertIntoBST(TreeNode root, int val) {
        TreeNode cur = root;
        while(cur != null) {
            if(val > cur.val) { // insert into right subtree
                if(cur.right == null) {
                    cur.right = new TreeNode(val);
                    return root;
                } else {
                    cur = cur.right;
                }
            } else {    // insert into left subtree
                if(cur.left == null) {
                    cur.left = new TreeNode(val);
                    return root;
                } else {
                    cur = cur.left;
                }
            }
        }
        return new TreeNode(val);
    }
}