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package class42;
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import java.util.LinkedList;
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import java.util.List;
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import java.util.Stack;
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public class Problem_0272_ClosestBinarySearchTreeValueII {
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public static class TreeNode {
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public int val;
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public TreeNode left;
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public TreeNode right;
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public TreeNode(int val) {
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this.val = val;
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}
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}
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// 这个解法来自讨论区的回答，最优解实现的很易懂且漂亮
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public static List<Integer> closestKValues(TreeNode root, double target, int k) {
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List<Integer> ret = new LinkedList<>();
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// >=8，最近的节点，而且需要快速找后继的这么一种结构
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Stack<TreeNode> moreTops = new Stack<>();
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// <=8，最近的节点，而且需要快速找前驱的这么一种结构
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Stack<TreeNode> lessTops = new Stack<>();
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getMoreTops(root, target, moreTops);
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getLessTops(root, target, lessTops);
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if (!moreTops.isEmpty() && !lessTops.isEmpty() && moreTops.peek().val == lessTops.peek().val) {
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getPredecessor(lessTops);
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}
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while (k-- > 0) {
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if (moreTops.isEmpty()) {
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ret.add(getPredecessor(lessTops));
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} else if (lessTops.isEmpty()) {
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ret.add(getSuccessor(moreTops));
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} else {
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double diffs = Math.abs((double) moreTops.peek().val - target);
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double diffp = Math.abs((double) lessTops.peek().val - target);
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if (diffs < diffp) {
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ret.add(getSuccessor(moreTops));
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} else {
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ret.add(getPredecessor(lessTops));
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}
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}
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}
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return ret;
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}
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// 在root为头的树上
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// 找到>=target，且最接近target的节点
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// 并且找的过程中，只要某个节点x往左走了，就把x放入moreTops里
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public static void getMoreTops(TreeNode root, double target, Stack<TreeNode> moreTops) {
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while (root != null) {
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if (root.val == target) {
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moreTops.push(root);
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break;
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} else if (root.val > target) {
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moreTops.push(root);
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root = root.left;
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} else {
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root = root.right;
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}
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}
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}
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// 在root为头的树上
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// 找到<=target，且最接近target的节点
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// 并且找的过程中，只要某个节点x往右走了，就把x放入lessTops里
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public static void getLessTops(TreeNode root, double target, Stack<TreeNode> lessTops) {
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while (root != null) {
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if (root.val == target) {
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lessTops.push(root);
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break;
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} else if (root.val < target) {
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lessTops.push(root);
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root = root.right;
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} else {
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root = root.left;
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}
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}
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}
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// 返回moreTops的头部的值
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// 并且调整moreTops : 为了以后能很快的找到返回节点的后继节点
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public static int getSuccessor(Stack<TreeNode> moreTops) {
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TreeNode cur = moreTops.pop();
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int ret = cur.val;
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cur = cur.right;
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while (cur != null) {
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moreTops.push(cur);
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cur = cur.left;
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}
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return ret;
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}
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// 返回lessTops的头部的值
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// 并且调整lessTops : 为了以后能很快的找到返回节点的前驱节点
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public static int getPredecessor(Stack<TreeNode> lessTops) {
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TreeNode cur = lessTops.pop();
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int ret = cur.val;
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cur = cur.left;
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while (cur != null) {
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lessTops.push(cur);
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cur = cur.right;
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}
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return ret;
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}
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}
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