Insert into a Binary Search Tree - LeetCode
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, 104].
-108 <= Node.val <= 108
All the values Node.val are unique.
-108 <= val <= 108
It's guaranteed that val does not exist in the original BST.
- code iteration
class Solution:
def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode:
node = root
while node:
if val > node.val:
if not node.right:
node.right = TreeNode(val)
return root
else:
node = node.right
else:
if not node.left:
node.left = TreeNode(val)
return root
else:
node = node.left
return TreeNode(val)
- code iteration
class Solution {
public TreeNode insertIntoBST(TreeNode root, int val) {
TreeNode res = root;
while (root != null){
if (root.val > val){
if (root.left != null) root = root.left;
else{
root.left = new TreeNode(val);
return res;
}
}
else if (root.val < val){
if (root.right != null) root = root.right;
else{
root.right = new TreeNode(val);
return res;
}
}
}
return new TreeNode(val);
}
}
- code recursion
class Solution:
def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if not root: return TreeNode(val)
if val < root.val:
root.left = self.insertIntoBST(root.left, val)
elif val > root.val:
root.right = self.insertIntoBST(root.right, val)
return root