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:

• 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.

 

 

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);
    }
}