Given the root node of a binary search tree (BST) and a value to be inserted into the tree, insert the value into the BST. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.
Note 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.
For example,
Given the tree:
4
/ \
2 7
/ \
1 3
And the value to insert: 5
You can return this binary search tree:
4
/ \
2 7
/ \ /
1 3 5
This tree is also valid:
5
/ \
2 7
/ \
1 3
\
4
Solution
We need to find node (based on BTS rules) that does not have left or right value (depends what value we are inserting) and it can be then appended into a tree.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode insertIntoBST(TreeNode root, int val) {
if (root == null) return null;
TreeNode current = root;
while (current != null) {
if (val < current.val) {
if (current.left == null) {
current.left = new TreeNode(val);
current = null;
} else {
current = current.left;
}
} else {
if (current.right == null) {
current.right = new TreeNode(val);
current = null;
} else {
current = current.right;
}
}
}
return root;
}
}
Recursive solution.
class Solution {
public TreeNode insertIntoBST(TreeNode root, int val) {
if (root == null) {
return new TreeNode(val); // return a new node if root is null
}
if (root.val < val) { // insert to the right subtree if val > root->val
root.right = insertIntoBST(root.right, val);
} else { // insert to the left subtree if val <= root->val
root.left = insertIntoBST(root.left, val);
}
return root;
}
}