The challenge
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, 10<sup>4</sup>]. -10<sup>8</sup> <= Node.val <= 10<sup>8</sup>- All the values
Node.valare unique. -10<sup>8</sup> <= val <= 10<sup>8</sup>- It’s guaranteed that
valdoes not exist in the original BST.
The solution in Java code
Our definition of a TreeNode is as follows:
/**
* 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;
* }
* }
*/
And our solution can be created as follows:
class Solution {
public TreeNode insertIntoBST(TreeNode root, int val) {
// if root node is null,
// then return a new TreeNode of the value
if (root==null) return new TreeNode(val);
// if the val is less than the root.val
if (val < root.val) {
//if left exists
if (root.left!=null)
this.insertIntoBST(root.left, val);
else {
// right exists
root.left = new TreeNode(val);
}
// val is not less than root.val
} else {
// if right exists
if (root.right!=null)
this.insertIntoBST(root.right, val);
else {
// left exists
root.right = new TreeNode(val);
}
}
// return the root node
return root;
}
}
