题目

给定一棵二叉搜索树和一个值k，请判断该二叉搜索树中是否存在值之和等于k的两个节点。假设二叉搜索树中节点的值均唯一。例如，在如图8.12所示的二叉搜索树中，存在值之和等于12的两个节点（节点5和节点7），但不存在值之和为22的两个节点。

分析1

解决这个问题最直观的思路是利用哈希表保存节点的值。可以采用任意遍历算法遍历输入的二叉搜索树，每遍历到一个节点（节点的值记为v），就在哈希表中查看是否存在值为k-v的节点。如果存在，就表示存在值之和等于k的两个节点。

解1

public class Test {public static void main(String[] args) {TreeNode node5 = new TreeNode(5);TreeNode node6 = new TreeNode(6);TreeNode node7 = new TreeNode(7);TreeNode node8 = new TreeNode(8);TreeNode node9 = new TreeNode(9);TreeNode node10 = new TreeNode(10);TreeNode node11 = new TreeNode(11);node8.left = node6;node8.right = node10;node6.left = node5;node6.right = node7;node10.left = node9;node10.right = node11;boolean result = findTarget(node8, 12);System.out.println(result);}public static boolean findTarget(TreeNode root, int k) {Set<Integer> set = new HashSet<>();Stack<TreeNode> stack = new Stack<>();TreeNode cur = root;while (cur != null || !stack.isEmpty()) {while (cur != null) {stack.push(cur);cur = cur.left;}cur = stack.pop();if (set.contains(k - cur.val)) {return true;}set.add(cur.val);cur = cur.right;}return false;}
}

分析2

面试题6介绍了如何利用双指针判断在排序数组中是否包含两个和为k的数字，即把第1个指针指向数组的第1个（也是最小的）数字，把第2个指针指向数组的最后一个（也是最大的）数字。如果两个数字之和等于k，那么就找到了两个符合要求的数字；如果两个数字之和大于k，那么向左移动第2个指针使它指向更小的数字；如果两个数字之和小于k，那么向右移动第1个指针使它指向更大的数字。

解2

public class Test {public static void main(String[] args) {TreeNode node5 = new TreeNode(5);TreeNode node6 = new TreeNode(6);TreeNode node7 = new TreeNode(7);TreeNode node8 = new TreeNode(8);TreeNode node9 = new TreeNode(9);TreeNode node10 = new TreeNode(10);TreeNode node11 = new TreeNode(11);node8.left = node6;node8.right = node10;node6.left = node5;node6.right = node7;node10.left = node9;node10.right = node11;boolean result = findTarget(node8, 12);System.out.println(result);}public static boolean findTarget(TreeNode root, int k) {if (root == null) {return false;}BSTIterator iterNext = new BSTIterator(root);BSTIteratorReversed iterPrev = new BSTIteratorReversed(root);int next = iterNext.next();int prev = iterPrev.prev();while (next != prev) {if (next + prev == k) {return true;}if (next + prev < k) {next = iterNext.next();}else {prev = iterPrev.prev();}}return false;}// 从小到大排列public static class BSTIterator {TreeNode cur;Stack<TreeNode> stack;public BSTIterator(TreeNode root) {cur = root;stack = new Stack<>();}public boolean hasNext() {return cur != null || !stack.isEmpty();}public int next() {while (cur != null) {stack.push(cur);cur = cur.left;}cur = stack.pop();int val = cur.val;cur = cur.right;return val;}}// 从大到小排列public static class BSTIteratorReversed {TreeNode cur;Stack<TreeNode> stack;public BSTIteratorReversed(TreeNode root) {cur = root;stack = new Stack<>();}public boolean hasPrev() {return cur != null || !stack.isEmpty();}public int prev() {while (cur != null) {stack.push(cur);cur = cur.right;}cur = stack.pop();int val = cur.val;cur = cur.left;return val;}}
}