【leetcode】701. Insert into a Binary Search Tree

题目如下：

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

解题思路：因为题目对插入后树的高度没有任何约束，所以最直接的方法就是对树进行遍历。如果遍历到的当前节点值大于给定值，判断其节点的左子节点是否存在：不存在则将给定值插入到其左子节点，存在则往左子节点继续遍历；如果小于，则同理，判断其右子节点。直到找到符合条件的节点为止。

代码如下：

# 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