題目如下：

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

解題思路：因為題目對插入後樹的高度沒有任何約束，所以最直接的方法就是對樹進行遍歷。如果遍歷到的當前節點值大於給定值，判斷其節點的左子節點是否存在：不存在則將給定值插入到其左子節點，存在則往左子節點繼續遍歷；如果小於，則同理，判斷其右子節點。直到找到符合條件的節點為止。

代碼如下：

# 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):
    loop = True
    def recursive(self,node,val):
        if self.loop == False:
            return
        if node.val > val:
            if node.left == None:
                node.left = TreeNode(val)
                self.loop = False
            else:
                self.recursive(node.left,val)
        else:
            if node.right == None:
                node.right = TreeNode(val)
                self.loop = False
            else:
                self.recursive(node.right,val)

    def insertIntoBST(self, root, val):
        """
        :type root: TreeNode
        :type val: int
        :rtype: TreeNode
        """
        if root == None:
            return TreeNode(val)
        self.loop = True
        self.recursive(root,val)
        return root
        

【leetcode】701. Insert into a Binary Search Tree