700. Search in a Binary Search Tree

最新推荐文章于 2024-09-01 11:32:18 发布

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最新推荐文章于 2024-09-01 11:32:18 发布

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分类专栏： LeetCode

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博客围绕二叉搜索树展开，提出需在给定的二叉搜索树中查找值等于给定值的节点，若存在则返回以该节点为根的子树，不存在则返回 NULL，还给出了相关示例，并提及了二叉树查找程序。

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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.

二叉树查找，程序如下所示：

/**
 * 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 root;
        }
        if (root.val == val){
            return root;
        }
        if (val >= root.val){
            return searchBST(root.right, val);
        }
        return searchBST(root.left, val);
    }
}