leetcode-cn Daily Challenge on September 30th, 2020.
leetcode Daily Challenge on October 6th, 2020.
Difficulty : Medium
Related Topics : Tree
Given the
root
node of a binary search tree (BST) and avalue
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.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.
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
- The number of nodes in the given tree will be between
0
and10^4
.- Each node will have a unique integer value from
0
to-10^8
, inclusive.-10^8 <= val <= 10^8
- It's guaranteed that
val
does not exist in the original BST.
- mine
- Java
Runtime: 0 ms, faster than 100.00%, Memory Usage: 39.6 MB, less than 98.74% of Java online submissions
public TreeNode insertIntoBST(TreeNode root, int val) { if(root == null) return new TreeNode(val); TreeNode t = root; if(t.val > val){ t.left = insertIntoBST(t.left, val); }else{ t.right = insertIntoBST(t.right, val); } return root; }
- Java