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 <= 108Node.val are unique.-108 <= val <= 108val does not exist in the original BST.
class Solution {
public TreeNode insertIntoBST(TreeNode root, int val) {
if(root == null){
TreeNode ans = new TreeNode(val);
return ans ;
}
if( root.val > val){
root.left = insertIntoBST(root.left , val);
}else{
root.right = insertIntoBST(root.right,val);
}
return root ;
}
}
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