树
二叉搜索树
二叉树
题目描述
给定二叉搜索树(BST)的根节点 root
和一个整数值 val
。
你需要在 BST 中找到节点值等于 val
的节点。 返回以该节点为根的子树。 如果节点不存在,则返回 null
。
示例 1:
输入: root = [4,2,7,1,3], val = 2
输出: [2,1,3]
示例 2:
输入: root = [4,2,7,1,3], val = 5
输出: []
提示:
树中节点数在 [1, 5000]
范围内
1 <= Node.val <= 107
root
是二叉搜索树
1 <= val <= 107
解法
方法一
Python3 Java C++ Go
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15 # 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 searchBST ( self , root : TreeNode , val : int ) -> TreeNode :
if root is None or root . val == val :
return root
return (
self . searchBST ( root . right , val )
if root . val < val
else self . searchBST ( root . left , val )
)
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23 /**
* 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 searchBST ( TreeNode root , int val ) {
if ( root == null || root . val == val ) {
return root ;
}
return root . val < val ? searchBST ( root . right , val ) : searchBST ( root . left , val );
}
}
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18 /**
* 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 * searchBST ( TreeNode * root , int val ) {
if ( ! root || root -> val == val ) return root ;
return root -> val < val ? searchBST ( root -> right , val ) : searchBST ( root -> left , val );
}
};
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17 /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func searchBST ( root * TreeNode , val int ) * TreeNode {
if root == nil || root . Val == val {
return root
}
if root . Val < val {
return searchBST ( root . Right , val )
}
return searchBST ( root . Left , val )
}
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