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701. Insert into a Binary Search Tree ​

题目 ​

You are given the root node of a binary search tree (BST) and a value to insert into the tree. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.

Notice 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.

 

Example 1:

Input: root = [4,2,7,1,3], val = 5
Output: [4,2,7,1,3,5]
Explanation: Another accepted tree is:

Example 2:

Input: root = [40,20,60,10,30,50,70], val = 25
Output: [40,20,60,10,30,50,70,null,null,25]

Example 3:

Input: root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
Output: [4,2,7,1,3,5]

 

Constraints:

  • The number of nodes in the tree will be in the range [0, 104].
  • -108 <= Node.val <= 108
  • All the values Node.val are unique.
  • -108 <= val <= 108
  • It's guaranteed that val does not exist in the original BST.
Related Topics
  • 树
  • 二叉搜索树
  • 二叉树

  • 👍 611
  • 👎 0

    • 解法 ​

    py
    # leetcode submit region begin(Prohibit modification and deletion)
# 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]:
        if not root:
            return TreeNode(val)

        if root.val > val: root.left = self.insertIntoBST(root.left, val)
        if root.val < val: root.right = self.insertIntoBST(root.right, val)

        return root

# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度 O(logN)
    • 空间复杂度 O(logN)