701. Insert into a Binary Search Tree

Problem 701

https://leetcode.com/problems/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,

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

Solution

這邊是以第一種（非重組）的 binary tree 來回傳 概念都是以 val 大於 root.val 就往 right node 小於就往 left.node 找，直到有 None 就換上 TreeNode(val)

loop

class SolutionII(object):
    def insertIntoBST(self, root, val):
        root_head = root
        root_val = TreeNode(val)
        while(True):
            if val < root.val:
                if root.left != None:
                    root = root.left
                else:
                    root.left = root_val
                    break
                    
            else:
                if root.right != None:
                    root = root.right
                else:
                    root.right = root_val
                    break
        return root_head

recursive

# 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):
        
        if root == None: return TreeNode(val)
        elif root.val < val : root.right = self.insertIntoBST( root.right, val )
        else: root.left = self.insertIntoBST( root.left, val )
        
        return root