LeetCode From Easy To Hard No9[Easy]: Search in a Binary Search Tree 查找二叉树中的值

查找二叉搜索树中特定值的子树

最新推荐文章于 2025-09-26 02:41:11 发布

原创最新推荐文章于 2025-09-26 02:41:11 发布 · 146 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#LeetCode #Easy

LeetCode 专栏收录该内容

11 篇文章

订阅专栏

本文介绍了一种算法，用于在二叉搜索树中查找具有特定值的节点，并返回以此节点为根的子树。通过递归方式实现，若找到目标值则返回节点，否则继续在左子树或右子树中查找。

Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node's value equals the given value. Return the subtree rooted with that node. If such node doesn't exist, you should return NULL.

For example,

Given the tree:
        4
       / \
      2   7
     / \
    1   3

And the value to search: 2

You should return this subtree:

      2     
     / \   
    1   3

In the example above, if we want to search the value 5, since there is no node with value 5, we should return NULL.

Note that an empty tree is represented by NULL, therefore you would see the expected output (serialized tree format) as [], not null.

查找二叉树中某个值，然后返回以该值为根节点的子树。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode searchBST(TreeNode root, int val) {
        if (root == null) return null;
        if (root.val == val) {
            return root;
        } else {
            // 二叉树的左子树值小于根节点值，所以如果根值大于搜索值，则递归遍历左树，否则递归右树
            return root.val > val ? searchBST(root.left, val) : searchBST(root.right, val);
        }
    }
}