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

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原创最新推荐文章于 2026-10-25 19:11:34 发布 168 阅读 AI 写同款· GEO 优化 ›

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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);
        }
    }
}

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