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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:
[0, 104]
.-108 <= Node.val <= 108
Node.val
are unique.-108 <= val <= 108
val
does not exist in the original BST.class Solution {
public TreeNode insertIntoBST(TreeNode root, int val) {
if(root == null) return new TreeNode(val);
if(root.val > val) root.left = insertIntoBST(root.left, val);
else root.right = insertIntoBST(root.right, val);
return root;
}
}
class Solution {
public:
TreeNode* insertIntoBST(TreeNode* root, int val) {
if(!root) return new TreeNode(val);
if(root->val > val) root->left = insertIntoBST(root->left, val);
else root->right = insertIntoBST(root->right, val);
return root;
}
};
class Solution:
def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if root is None: return TreeNode(val)
if root.val > val: root.left = self.insertIntoBST(root.left, val)
else: root.right = self.insertIntoBST(root.right, val)
return root
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