701 Insert into a Binary Search Tree 二叉搜索树中的插入操作
Description:
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:
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, 10^4].
-10^8 <= Node.val <= 10^8
All the values Node.val are unique.
-10^8 <= val <= 10^8
It's guaranteed that val does not exist in the original BST.
题目描述:
给定二叉搜索树（BST）的根节点和要插入树中的值，将值插入二叉搜索树。 返回插入后二叉搜索树的根节点。 输入数据 保证 ，新值和原始二叉搜索树中的任意节点值都不同。
注意，可能存在多种有效的插入方式，只要树在插入后仍保持为二叉搜索树即可。 你可以返回 任意有效的结果 。
示例 :
示例 1：
输入：root = [4,2,7,1,3], val = 5
输出：[4,2,7,1,3,5]
解释：另一个满足题目要求可以通过的树是：
示例 2：
输入：root = [40,20,60,10,30,50,70], val = 25
输出：[40,20,60,10,30,50,70,null,null,25]
示例 3：
输入：root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
输出：[4,2,7,1,3,5]
提示:
给定的树上的节点数介于 0 和 10^4 之间
每个节点都有一个唯一整数值，取值范围从 0 到 10^8
-10^8 <= val <= 10^8
新值和原始二叉搜索树中的任意节点值都不同
思路:
搜索到能插入的叶子节点即可
时间复杂度为平均 O(lgn), 最差为 O(n), 空间复杂度为 O(1)
代码:
C++:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution
{
public:
TreeNode* insertIntoBST(TreeNode* root, int val)
{
if (!root) return new TreeNode(val);
TreeNode *parent = root, *p = root;
while (p)
{
parent = p;
p = p -> val < val ? p -> right : p -> left;
}
if (parent -> val < val) parent -> right = new TreeNode(val);
else parent -> left = new TreeNode(val);
return root;
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode insertIntoBST(TreeNode root, int val) {
if (root == null) return new TreeNode(val);
if (root.val < val) root.right = insertIntoBST(root.right, val);
else root.left = insertIntoBST(root.left, val);
return root;
}
}
Python:
# 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: TreeNode, val: int) -> TreeNode:
if root is None:
return TreeNode(val)
if root.val < val:
root.right = self.insertIntoBST(root.right, val)
else:
root.left = self.insertIntoBST(root.left, val)
return root
