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 = 9Output: true

Example 2:

Input: root = [5,3,6,2,4,null,7], k = 28Output: false

Example 3:

Input: root = [2,1,3], k = 4Output: true

Example 4:

Input: root = [2,1,3], k = 1Output: 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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