LeetCode Easy653中两数之和输入为BST的示例分析
LeetCode Easy653中两数之和输入为BST的示例分析,针对这个问题,这篇文章详细介绍了相对应的分析和解答,希望可以帮助更多想解决这个问题的小伙伴找到更简单易行的方法。
Description
https://leetcode.com/problems/two-sum-iv-input-is-a-bst/
Given the root of a Binary Search Tree and a target number k, return true if there exist two elements in the BST such that their sum is equal to the given target.
Example 1:
Input: root = [5,3,6,2,4,null,7], k = 9 Output: true
Example 2:
Input: root = [5,3,6,2,4,null,7], k = 28 Output: false
Example 3:
Input: root = [2,1,3], k = 4 Output: true
Example 4:
Input: root = [2,1,3], k = 1 Output: false
Example 5:
Input: root = [2,1,3], k = 3 Output: true
Constraints:
The number of nodes in the tree is in the range $[1, 10^4]$.
$-10^4 <= Node.val <= 10^4$
root
is guaranteed to be a valid binary search tree.$-10^5 <= k <= 10^5$
Analysis
用两指针方法,一指针前序遍历,另一指针二分查找。
Submission
public class TwoSumIVInputIsABST { public boolean findTarget(TreeNode root, int k) { return traverse(root, root, k); } // 前序遍历 private boolean traverse(TreeNode node, TreeNode root, int k) { if (node == null) return false; // node.val恰好是k的一半时,根据BST特性,没必要找另一个节点 if (2 * node.val != k && findTarget2(root, k - node.val)) return true; return traverse(node.left, root, k) || traverse(node.right, root, k); } // 二分查找 private boolean findTarget2(TreeNode node, int targetValue) { if (node == null) return false; if (node.val == targetValue) return true; return findTarget2(targetValue < node.val ? node.left : node.right, targetValue); } }
Test
import static org.junit.Assert.*; import org.junit.Test; import com.lun.util.BinaryTree; import com.lun.util.BinaryTree.TreeNode; public class TwoSumIVInputIsABSTTest { @Test public void test() { TwoSumIVInputIsABST obj = new TwoSumIVInputIsABST(); TreeNode root1 = BinaryTree.integers2BinaryTree(5, 3, 6, 2, 4, null, 7); assertTrue(obj.findTarget(root1, 9)); assertFalse(obj.findTarget(root1, 28)); TreeNode root2 = BinaryTree.integers2BinaryTree(2, 1, 3); assertTrue(obj.findTarget(root2, 4)); assertFalse(obj.findTarget(root2, 1)); assertTrue(obj.findTarget(root2, 3)); } }
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