[LeetCode] 701. Insert into a Binary Search Tree_Medium_tag: 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, 

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

这个题目利用recursive的方式，去判断是大于root.val 还是小于，因为题目说了不会有重复的，所以类似于在BST中找到node，去分别recursive得到left 和right child，最后返回root。

08/17/2023 Update, 利用iterable 的方式


Code

class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

class Solution:
    def insertBST(self, root, val):
        if not root: return TreeNode(val)
        if root.val < val:
            root.right = self.insertBST(root.right, val)
        else:
            root.left = self.insertBST(root.left, val)
        return root

 

Code 2

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
        add_node = TreeNode(val)
        node = root
        while node:
            if node.val < val:
                if node.right:
                    node = node.right
                else:
                    node.right = add_node
                    return root
            else:
                if node.left:
                    node = node.left
                else:
                    node.left = add_node
                    return root
        return add_node