700. Search in a Binary Search Tree 在bst中返回目标所在的最小子树
You are given the root
of a binary search tree (BST) and an integer val
.
Find the node in the BST that the node's value equals val
and return the subtree rooted with that node. If such a node does not exist, return null
.
Example 1:
Input: root = [4,2,7,1,3], val = 2
Output: [2,1,3]
Example 2:
Input: root = [4,2,7,1,3], val = 5 Output: []
不懂的地方:
不懂怎么才能返回的是一棵树:自然而然形成一棵树。因为没有继续递归下去了,所以就还带着附属的左右节点。
有时候low/high只要有一个target就行了,所以其实也是traverse模板的变形!
class Solution {
public TreeNode searchBST(TreeNode root, int val) {
if (root == null)
return root;
if (root.val == val)
return root;
else if (root.val > val)
return searchBST(root.left, val);
else if (root.val < val)
return searchBST(root.right, val);
return null;
}
}