# 701. Insert into a Binary Search Tree



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,&#x20;

```
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
```

Recursive

```python
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def insertIntoBST(self, root, val):
        """
        :type root: TreeNode
        :type val: int
        :rtype: TreeNode
        """
        if not root:
            return TreeNode(val)
        if val < root.val:
            root.left = self.insertIntoBST(root.left, val)
        else:
            root.right = self.insertIntoBST(root.right, val)
        return root
```

Iterative

```python
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def insertIntoBST(self, root, val):
        """
        :type root: TreeNode
        :type val: int
        :rtype: TreeNode
        """
        if not root:
            return TreeNode(val)
        pre = None
        cur = root
        while(cur):
            pre = cur
            if val < cur.val:
                cur = cur.left
            else:
                cur = cur.right
        if val < pre.val:
            pre.left = TreeNode(val)
        else:
            pre.right = TreeNode(val)
        return root
```
