題目

https://leetcode.com/problems/serialize-and-deserialize-bst/

題解

本題的難點(diǎn)在于 利用 BST 的性質(zhì)。

幾個(gè)提示

• 根據(jù)后序遍歷BST性質(zhì)，頭結(jié)點(diǎn)是最后一個(gè)元素，且有一個(gè)分界，滿足比頭結(jié)點(diǎn)小的都在左邊，比頭結(jié)點(diǎn)大的都在右邊
• 本題滿足條件（但是題目中沒說）沒有重復(fù)值
• 要求你返回的樹的形狀不僅是 BST，而且要與輸入完全相同。同樣是 BST，注意以下兩樹的區(qū)別。它們中序序列相同，后序序列不同。
  

對比

一開始我想，這不就是序列化/反序列化一棵普通樹嗎？只不過給你的時(shí)候恰好是 BST。這和 297. Serialize and Deserialize Binary Tree 有什么不同？

再一想，不一樣。利用二叉樹的性質(zhì)，可以將序列化結(jié)果簡化，也就是本題說的 The encoded string should be as compact as possible. 編碼的字符串應(yīng)盡可能緊湊。

參考左神算法書 P107 “二叉樹的序列化和反序列化”，他在序列化一棵普通樹的時(shí)候，利用了先序遍歷的方式，用 # 記錄空節(jié)點(diǎn)。而本題是 BST，其本身性質(zhì)決定了不需要通過 # 記錄空節(jié)點(diǎn)，也能進(jìn)行 唯一結(jié)果的 序列化與反序列化。

附從錯(cuò)誤思路到正確思路的草稿：

/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode(int x) { val = x; }* }*/ class Codec {// Encodes a tree to a single string.public String serialize(TreeNode root) {if (root == null) return "";StringBuilder s = new StringBuilder();postOrder(root, s);return s.substring(0, s.length() - 1); // 移除末尾 ','}public void postOrder(TreeNode node, StringBuilder s) {if (node == null) return;postOrder(node.left, s);postOrder(node.right, s);s.append(node.val).append(",");}// Decodes your encoded data to tree.public TreeNode deserialize(String data) {if (data.equals("")) return null;String[] ss = data.split(",");int[] arr = new int[ss.length];for (int i = 0; i < ss.length; i++) {arr[i] = Integer.parseInt(ss[i]);}return buildBST(arr, 0, arr.length - 1);}// 根據(jù)后序遍歷BST性質(zhì)，頭結(jié)點(diǎn)是最后一個(gè)元素，且有一個(gè)分界，滿足比頭結(jié)點(diǎn)小的都在左邊，比頭結(jié)點(diǎn)大的都在右邊public TreeNode buildBST(int[] arr, int L, int R) {if (L > R) return null;TreeNode node = new TreeNode(arr[R]);// 找左右子樹的分界點(diǎn)int M = L;while (arr[M] < arr[R]) {M++;}// M指向分界點(diǎn)右邊第一個(gè)元素，本題缺隱含條件：沒有重復(fù)值！node.left = buildBST(arr, L, M - 1);node.right = buildBST(arr, M, R - 1);return node;} }// Your Codec object will be instantiated and called as such: // Codec ser = new Codec(); // Codec deser = new Codec(); // String tree = ser.serialize(root); // TreeNode ans = deser.deserialize(tree); // return ans;

可參考：左程云《程序員代碼面試指南》根據(jù)后序數(shù)組重建搜索二叉樹

package chapter_3_binarytreeproblem;public class Problem_14_PosArrayToBST {public static boolean isPostArray(int[] arr) {if (arr == null || arr.length == 0) {return false;}return isPost(arr, 0, arr.length - 1);}public static boolean isPost(int[] arr, int start, int end) {if (start == end) {return true;}int less = -1;int more = end;for (int i = start; i < end; i++) {if (arr[end] > arr[i]) {less = i;} else {more = more == end ? i : more;}}if (less == -1 || more == end) {return isPost(arr, start, end - 1);}if (less != more - 1) {return false;}return isPost(arr, start, less) && isPost(arr, more, end - 1);}public static class Node {public int value;public Node left;public Node right;public Node(int value) {this.value = value;}}public static Node posArrayToBST(int[] posArr) {if (posArr == null) {return null;}return posToBST(posArr, 0, posArr.length - 1);}public static Node posToBST(int[] posArr, int start, int end) {if (start > end) {return null;}Node head = new Node(posArr[end]);int less = -1;int more = end;for (int i = start; i < end; i++) {if (posArr[end] > posArr[i]) {less = i;} else {more = more == end ? i : more;}}head.left = posToBST(posArr, start, less);head.right = posToBST(posArr, more, end - 1);return head;}// for test -- print treepublic static void printTree(Node head) {System.out.println("Binary Tree:");printInOrder(head, 0, "H", 17);System.out.println();}public static void printInOrder(Node head, int height, String to, int len) {if (head == null) {return;}printInOrder(head.right, height + 1, "v", len);String val = to + head.value + to;int lenM = val.length();int lenL = (len - lenM) / 2;int lenR = len - lenM - lenL;val = getSpace(lenL) + val + getSpace(lenR);System.out.println(getSpace(height * len) + val);printInOrder(head.left, height + 1, "^", len);}public static String getSpace(int num) {String space = " ";StringBuffer buf = new StringBuffer("");for (int i = 0; i < num; i++) {buf.append(space);}return buf.toString();}public static void main(String[] args) {int[] arr = { 2, 1, 3, 6, 5, 7, 4 };System.out.println(isPost(arr, 0, arr.length - 1));printTree(posArrayToBST(arr));} }

總結(jié)

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