给定二叉搜索树(BST)的根节点和一个值。你需要在BST中找到节点值等于给定值的节点。返回以该节点为根的子树。如果节点不存在,则返回 NULL。
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.
例如,
给定二叉搜索树:
4
/ \
2 7
/ \
1 3
和值: 2
你应该返回如下子树:
2
/ \
1 3
在上述示例中,如果要找的值是 5
,但因为没有节点值为 5
,我们应该返回 NULL
。
二叉搜索树中的搜索操作, 可根据 BST 的特性,对于每个节点:
Java:
class Solution {
public TreeNode searchBST(TreeNode root, int val) {
if (root == null)
return null;
if (val == root.val)
return root;
if (val > root.val)
return searchBST(root.right, val);
else
return searchBST(root.left, val);
}
}
Python:
class Solution:
def searchBST(self, root: TreeNode, val: int) -> TreeNode:
if not root:
return None
if val == root.val:
return root
if val > root.val:
return self.searchBST(root.right, val)
else:
return self.searchBST(root.left, val)
Golang:
func searchBST(root *TreeNode, val int) *TreeNode {
if root == nil {
return nil
}
if root.Val == val {
return root
}
if val > root.Val {
return searchBST(root.Right, val)
} else {
return searchBST(root.Left, val)
}
}
Java:
class Solution {
public TreeNode searchBST(TreeNode root, int val) {
while(root!=null){
if (root.val == val)
return root;
if (val > root.val)
root=root.right;
else
root=root.left;
}
return null;
}
}
Python:
class Solution:
def searchBST(self, root: TreeNode, val: int) -> TreeNode:
while root:
if root.val == val:
return root
if val > root.val:
root = root.right
else:
root = root.left
return None
Golang:
func searchBST(root *TreeNode, val int) *TreeNode {
for root != nil {
if root.Val == val {
return root
}
if val > root.Val {
root = root.Right
} else {
root = root.Left
}
}
return nil
}