我今天去参加一个面试，我被要求序列化一棵二叉树。我实现了一种基于数组的方法，其中节点 i 的子节点（按级别顺序遍历编号）位于左子节点的 2*i 索引处，右子节点位于 2*i + 1 处。面试官似乎或多或少很高兴，但我想知道序列化到底是什么意思？它是否特别涉及扁平化树以写入磁盘，或者序列化树是否还包括将树转换为链表，例如。另外，我们如何将树扁平化为（双）链表，然后重建它？你能从链表中重新创建树的确切结构吗？
问问题
41854 次
11 回答
13
所有这些文章都主要讨论序列化部分。反序列化部分一次性完成有点棘手。
我也为反序列化实施了一个有效的解决方案。
问题：序列化和反序列化包含正数的二叉树。
序列化部分：
- 使用 0 表示空值。
- 使用前序遍历序列化为整数列表。
反序列化部分：
- 接受整数列表并使用递归辅助方法进行反序列化。
- 递归反序列化器返回一对 (BTNode node, int nextIndexToRead)，其中 node 是到目前为止构造的树节点， nextIndexToRead 是序列化数字列表中要读取的下一个数字的位置。
下面是Java中的代码：
public final class BinaryTreeSerializer
{
public static List<Integer> Serialize(BTNode root)
{
List<Integer> serializedNums = new ArrayList<Integer>();
SerializeRecursively(root, serializedNums);
return serializedNums;
}
private static void SerializeRecursively(BTNode node, List<Integer> nums)
{
if (node == null)
{
nums.add(0);
return;
}
nums.add(node.data);
SerializeRecursively(node.left, nums);
SerializeRecursively(node.right, nums);
}
public static BTNode Deserialize(List<Integer> serializedNums)
{
Pair pair = DeserializeRecursively(serializedNums, 0);
return pair.node;
}
private static Pair DeserializeRecursively(List<Integer> serializedNums, int start)
{
int num = serializedNums.get(start);
if (num == 0)
{
return new Pair(null, start + 1);
}
BTNode node = new BTNode(num);
Pair p1 = DeserializeRecursively(serializedNums, start + 1);
node.left = p1.node;
Pair p2 = DeserializeRecursively(serializedNums, p1.startIndex);
node.right = p2.node;
return new Pair(node, p2.startIndex);
}
private static final class Pair
{
BTNode node;
int startIndex;
private Pair(BTNode node, int index)
{
this.node = node;
this.startIndex = index;
}
}
}
public class BTNode
{
public int data;
public BTNode left;
public BTNode right;
public BTNode(int data)
{
this.data = data;
}
}
于 2013-08-24T20:28:43.597 回答
9
使用前序遍历，序列化二叉树。使用相同的前序遍历来反序列化树。小心边缘情况。这里空节点用“#”表示
public static String serialize(TreeNode root){
StringBuilder sb = new StringBuilder();
serialize(root, sb);
return sb.toString();
}
private static void serialize(TreeNode node, StringBuilder sb){
if (node == null) {
sb.append("# ");
} else {
sb.append(node.val + " ");
serialize(node.left, sb);
serialize(node.right, sb);
}
}
public static TreeNode deserialize(String s){
if (s == null || s.length() == 0) return null;
StringTokenizer st = new StringTokenizer(s, " ");
return deserialize(st);
}
private static TreeNode deserialize(StringTokenizer st){
if (!st.hasMoreTokens())
return null;
String val = st.nextToken();
if (val.equals("#"))
return null;
TreeNode root = new TreeNode(Integer.parseInt(val));
root.left = deserialize(st);
root.right = deserialize(st);
return root;
}
于 2016-01-18T17:56:33.533 回答
7
方法 1：同时进行 Inorder 和 Preorder 遍历以对树数据进行序列化。在反序列化时使用 Pre-order 并在 In order 上执行 BST 以正确形成树。
您需要两者，因为 A -> B -> C 可以表示为预购，即使结构可以不同。
方法 2：使用 # 作为哨兵，无论左孩子或右孩子为空......
于 2013-02-23T19:49:53.920 回答
1
我一直在试图了解它的要点。所以这是我的Java实现。如前所述，这是一棵二叉树而不是 BST。对于序列化，前序遍历似乎更容易（对于空节点的“NULL”字符串）。请使用递归调用的完整示例检查下面的代码。对于反序列化，字符串被转换为 LinkedList，其中 remove(0) 获取 O(1) 运行时间内的顶部元素。另请参阅反序列化代码注释中的完整示例。希望这能帮助比我更努力的人:) 每种方法（序列化和反序列化）的总体运行时间与二叉树遍历的运行时间相同，即 O(n)，其中 n 是节点数（条目）在树上
import java.util.LinkedList;
import java.util.List;
public class SerDesBinTree<T> {
public static class TreeEntry<T>{
T element;
TreeEntry<T> left;
TreeEntry<T> right;
public TreeEntry(T x){
element = x;
left = null;
right = null;
}
}
TreeEntry<T> root;
int size;
StringBuilder serSB = new StringBuilder();
List<String> desList = new LinkedList<>();
public SerDesBinTree(){
root = null;
size = 0;
}
public void traverseInOrder(){
traverseInOrder(this.root);
}
public void traverseInOrder(TreeEntry<T> node){
if (node != null){
traverseInOrder(node.left);
System.out.println(node.element);
traverseInOrder(node.right);
}
}
public void serialize(){
serialize(this.root);
}
/*
* 1
* / \
* 2 3
* /
* 4
*
* ser(1)
* serSB.append(1) serSB: 1
* ser(1.left)
* ser(1.right)
* |
* |
* ser(1.left=2)
* serSB.append(2) serSB: 1, 2
* ser(2.left)
* ser(2.right)
* |
* |
* ser(2.left=null)
* serSB.append(NULL) serSB: 1, 2, NULL
* return
* |
* ser(2.right=null)
* serSB.append(NULL) serSB: 1, 2, NULL, NULL
* return
*
* |
* ser(1.right=3)
* serSB.append(3) serSB: 1, 2, NULL, NULL, 3
* ser(3.left)
* ser(3.right)
*
* |
* ser(3.left=4)
* serSB.append(4) serSB: 1, 2, NULL, NULL, 3, 4
* ser(4.left)
* ser(4.right)
*
* |
* ser(4.left=null)
* serSB.append(NULL) serSB: 1, 2, NULL, NULL, 3, 4, NULL
* return
*
* ser(4.right=null)
* serSB.append(NULL) serSB: 1, 2, NULL, NULL, 3, 4, NULL, NULL
* return
*
* ser(3.right=null)
* serSB.append(NULL) serSB: 1, 2, NULL, NULL, 3, 4, NULL, NULL, NULL
* return
*
*/
public void serialize(TreeEntry<T> node){
// preorder traversal to build the string
// in addition: NULL will be added (to make deserialize easy)
// using StringBuilder to append O(1) as opposed to
// String which is immutable O(n)
if (node == null){
serSB.append("NULL,");
return;
}
serSB.append(node.element + ",");
serialize(node.left);
serialize(node.right);
}
public TreeEntry<T> deserialize(TreeEntry<T> newRoot){
// convert the StringBuilder into a list
// so we can do list.remove() for the first element in O(1) time
String[] desArr = serSB.toString().split(",");
for (String s : desArr){
desList.add(s);
}
return deserialize(newRoot, desList);
}
/*
* 1
* / \
* 2 3
* /
* 4
*
* deser(root, list) list: 1, 2, NULL, NULL, 3, 4, NULL, NULL, NULL
* root = new TreeEntry(1) list: 2, NULL, NULL, 3, 4, NULL, NULL, NULL
* root.left = deser(root.left, list) // **
* root.right = deser(root.right, list) // *-*
* return root // ^*^
*
*
* so far subtree
* 1
* / \
* null null
*
* deser(root.left, list)
* root.left = new TreeEntry(2) list: NULL, NULL, 3, 4, NULL, NULL, NULL
* root.left.left = deser(root.left.left, list) // ***
* root.left.right = deser(root.left.right, list) // ****
* return root.left // eventually return new TreeEntry(2) to ** above after the two calls are done
*
* so far subtree
* 2
* / \
* null null
*
* deser(root.left.left, list)
* // won't go further down as the next in list is NULL
* return null // to *** list: NULL, 3, 4, NULL, NULL, NULL
*
* so far subtree (same, just replacing null)
* 2
* / \
* null null
*
* deser(root.left.right, list)
* // won't go further down as the next in list is NULL
* return null // to **** list: 3, 4, NULL, NULL, NULL
*
* so far subtree (same, just replacing null)
* 2
* / \
* null null
*
*
* so far subtree // as node 2 completely returns to ** above
* 1
* / \
* 2 null
* / \
* null null
*
*
* deser(root.right, list)
* root.right = new TreeEntry(3) list: 4, NULL, NULL, NULL
* root.right.left = deser(root.right.left, list) // *&*
* root.right.right = deser(root.right.right, list) // *---*
* return root.right // eventually return to *-* above after the previous two calls are done
*
* so far subtree
* 3
* / \
* null null
*
*
* deser(root.right.left, list)
* root.right.left = new TreeEntry(4) list: NULL, NULL, NULL
* root.right.left.left = deser(root.right.left.left, list) // *(*
* root.right.left.right = deser(root.right.left.right, list) // *)*
* return root.right.left // to *&*
*
* so far subtree
* 4
* / \
* null null
*
* deser(root.right.left.left, list)
* // won't go further down as the next in list is NULL
* return null // to *(* list: NULL, NULL
*
* so far subtree (same, just replacing null)
* 4
* / \
* null null
*
* deser(root.right.left.right, list)
* // won't go further down as the next in list is NULL
* return null // to *)* list: NULL
*
*
* so far subtree (same, just replacing null)
* 4
* / \
* null null
*
*
* so far subtree
* 3
* / \
* 4 null
* / \
* null null
*
*
* deser(root.right.right, list)
* // won't go further down as the next in list is NULL
* return null // to *---* list: empty
*
* so far subtree (same, just replacing null of the 3 right)
* 3
* / \
* 4 null
* / \
* null null
*
*
* now returning the subtree rooted at 3 to root.right in *-*
*
* 1
* / \
* / \
* / \
* 2 3
* / \ / \
* null null / null
* /
* 4
* / \
* null null
*
*
* finally, return root (the tree rooted at 1) // see ^*^ above
*
*/
public TreeEntry<T> deserialize(TreeEntry<T> node, List<String> desList){
if (desList.size() == 0){
return null;
}
String s = desList.remove(0); // efficient operation O(1)
if (s.equals("NULL")){
return null;
}
Integer sInt = Integer.parseInt(s);
node = new TreeEntry<T>((T)sInt);
node.left = deserialize(node.left, desList);
node.right = deserialize(node.right, desList);
return node;
}
public static void main(String[] args) {
/*
* 1
* / \
* 2 3
* /
* 4
*
*/
SerDesBinTree<Integer> tree = new SerDesBinTree<>();
tree.root = new TreeEntry<Integer>(1);
tree.root.left = new TreeEntry<Integer>(2);
tree.root.right = new TreeEntry<Integer>(3);
tree.root.right.left = new TreeEntry<Integer>(4);
//tree.traverseInOrder();
tree.serialize();
//System.out.println(tree.serSB);
tree.root = null;
//tree.traverseInOrder();
tree.root = tree.deserialize(tree.root);
//tree.traverseInOrder();
// deserialize into a new tree
SerDesBinTree<Integer> newTree = new SerDesBinTree<>();
newTree.root = tree.deserialize(newTree.root);
newTree.traverseInOrder();
}
}
于 2017-12-21T16:43:23.330 回答
0
如何执行按顺序遍历并将根键和所有节点键放入 std::list 或您选择的其他容器中，以使树变平。然后，只需使用 boost 库序列化您选择的 std::list 或容器。
反过来很简单，然后使用标准插入二叉树来重建树。对于非常大的树，这可能并不完全有效，但是将树转换为 std::list 的运行时间最多为 O(n)，而重建树的时间最多为 O(log n)。
当我将我的数据库从 Java 转换为 C++ 时，我将要序列化我刚刚用 C++ 编写的树。
于 2013-03-12T17:59:28.297 回答
0
最好的方法是使用一个特殊的字符（如前面提到的 # ）作为哨兵。它比构造一个中序遍历数组和一个前序/后序遍历数组要好，无论是在空间复杂度方面还是时间复杂度方面。它也更容易实现。
链表在这里不太合适，因为为了重建树，你最好有 const 元素访问时间
于 2014-03-19T22:07:29.500 回答
0
我没有使用预购，但我使用的是 BFS。这是 leetcode的一个问题
大多数人在使用预购时执行不正确：预期结果应该是
“[1,2,3,null,null,4,5]”，但大多数人将输出打印为“[1,2,3,null,null,4,5,null,null]”，因为它们是不计算级别。
这是我的实现，结果正确。
class Node(object):
def __init__(self,data):
self.left = None
self.right = None
self.data = data
def serialize(root):
queue = [(root,0)]
result = []
max_level_with_value = 0
while queue:
(node,l) = queue.pop(0)
if node:
result.append((node.data,l))
queue.extend([(node.left,l+1),
(node.right,l+1)
])
max_level_with_value = max(max_level_with_value,l)
else:
result.append(('null',l))
filter_redundant(result,max_level_with_value)
def filter_redundant(result,max_level_with_value):
for v,l in result:
if l<= max_level_with_value:
print(v)
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.right.left = Node(4)
root.right.right = Node(5)
serialize(root)
于 2019-10-08T23:02:14.723 回答
0
序列化反序列化二叉树：
public class BTSerialization {
public static class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
// Encodes a tree to a single string.
public String serialize(TreeNode root) {
StringBuilder sb = new StringBuilder();
serialize(root, sb);
return sb.toString();
}
private void serialize(TreeNode root, StringBuilder sb) {
if (root == null) {
sb.append("()");
return;
}
sb.append('(').append(root.val);
serialize(root.left, sb);
serialize(root.right, sb);
sb.append(')');
}
// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
if ("()".equals(data)) return null;
data = data.substring(1, data.length() - 1); // Unwrap the two parenthesis (root(left)(right))
int i = 0;
while (data.charAt(i) != '(') i++;
TreeNode root = new TreeNode(Integer.parseInt(data.substring(0, i)));
data = data.substring(i);
i = -1;
int open = 0;
while (true) { // Find left child- starts with parenthesis
i++;
if (data.charAt(i) == '(') open++;
else if (data.charAt(i) == ')') {
open--;
if (open == 0) break;
}
}
root.left = deserialize(data.substring(0, i + 1));
data = data.substring(i + 1); // The rest is right child- starts with parenthesis
root.right = deserialize(data);
return root;
}
public static void main(String[] args) {
BTSerialization b = new BTSerialization();
TreeNode node1 = new TreeNode(1);
TreeNode node2 = new TreeNode(2);
TreeNode node3 = new TreeNode(3);
TreeNode node4 = new TreeNode(4);
TreeNode node5 = new TreeNode(5);
node1.left = node2;
node1.right = node3;
node3.left = node4;
node3.right = node5;
TreeNode root = b.deserialize(b.serialize(node1));
System.out.println();
}
}
序列化反序列化 N 叉树： 使用此代码，您可以序列化、反序列化任何 N 叉树。基本上，二叉树是二叉树。N 表示整棵树中所有节点的最大子节点数。
public class CodecNry {
class Node {
public int val;
public List<Node> children;
public Node(int _val, List<Node> _children) {
val = _val;
children = _children;
}
}
public String serialize(Node root) {
if (root == null) return ""; // Serialization base case
StringBuilder stringBuilder = new StringBuilder();
stringBuilder.append(root.val).append(",").append(root.children.size()); // Add root val+,+num of children
for (Node child : root.children)
stringBuilder.append(",").append(serialize(child)); // Add children recursively, one by one
return stringBuilder.toString(); // Return result
}
int i;
public Node deserialize(String data) {
if (data.isBlank()) return null; // Base case root is null
i = 0; // The index on the tokens
return recursiveDeserialize(data.split(",")); // Recursively build the tree from the tokenized string
}
private Node recursiveDeserialize(String[] nums) {
if (i == nums.length) return null; // Base case- no more child
Node root = new Node(Integer.parseInt(nums[i++]), new ArrayList<>()); // Build the root
int childrenCount = Integer.parseInt(nums[i++]); // Get number of children
for (int j = 0; j < childrenCount; j++) { // Add children recursively one by one
Node child = recursiveDeserialize(nums);
if (child != null) root.children.add(child); // If child is not null, add it to the children of root
}
return root; // Return the root of the tree
}
}
于 2021-03-03T02:38:17.157 回答
0
这是 Python 中的一个迟到的答案。它使用（深度优先）预序序列化并返回strings. 反序列化返回树。
class Node:
def __init__(self, val, left=None, right=None):
self.val = val
self.left = left
self.right = right
# This method serializes the tree into a string
def serialize(root):
vals = []
def encode(node):
vals.append(str(node.val))
if node.left is not None:
encode(node.left)
else:
vals.append("L")
if node.right is not None:
encode(node.right)
else:
vals.append("R")
encode(root)
print(vals)
return vals
# This method deserializes the string back into the tree
def deserialize(string_list):
def create_a_tree(sub_list):
if sub_list[0] == 'L' or sub_list[0] == 'R':
del sub_list[0]
return None
parent = Node(sub_list[0])
del sub_list[0]
parent.left = create_a_tree(sub_list)
parent.right = create_a_tree(sub_list)
return parent
if len(string_list) != 0:
root_node = create_a_tree(string_list)
else:
print("ERROR - empty string!")
return 0
return root_node
去测试：
tree1 = Node('root', Node('left'), Node('right'))
t = deserialize(serialize(tree1))
print(str(t.right.val))
于 2019-02-28T02:19:34.550 回答
0
将其转换为大小为 (2n + 1) 的数组，每个左子和右子将相应地代替 (2 * 节点号) 和 ((2 * 节点号) + 1。
于 2020-11-05T12:37:19.390 回答
-2
序列化是将数据结构或对象转换为位序列的过程，以便将其存储在文件或内存缓冲区中，或通过网络连接链路传输，以便稍后在相同或另一个计算机环境中重建。
反序列化是将字符串转换回原始树结构的过程。
序列化和反序列化的概念与编译器对代码所做的非常相似。整个编译过程有多个阶段，但我们会尽量保持抽象。
给定一段代码，编译器将不同的定义良好的组件分解为标记（例如，int 是一个标记，double 是另一个标记，{ 是一个标记，} 是另一个标记，等等）。[链接到编译的抽象级别的演示][1]。
序列化：我们使用前序遍历逻辑将树序列化为字符串。我们将添加“X”来表示树中的空指针/节点。此外，为了牢记我们的反序列化逻辑，我们需要在每个序列化的节点值之后添加“，”，以便反序列化过程可以访问用“，”分割的每个节点值。
Leetcode 链接：https ://leetcode.com/problems/serialize-and-deserialize-binary-tree/
Back To Back SWE Youtube 频道的解释：https ://www.youtube.com/watch?v=suj1ro8TIVY
For example:
You may serialize the following tree:
1
/ \
2 3
/ \
4 5
as "[1,2,null,null,3,4,null,null,5,null,null,]"
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Codec {
// Encodes a tree to a single string.
public String serialize(TreeNode root) {
if(root == null)
return "X,";
String leftSerialized = serialize(root.left);
String rightSerialized = serialize(root.right);
return root.val + "," + leftSerialized + rightSerialized;
}
private TreeNode deserializeHelper(Queue<String> queue)
{
String nodeValue = queue.poll();
if(nodeValue.equals("X"))
return null;
TreeNode newNode = new TreeNode(Integer.valueOf(nodeValue));
newNode.left = deserializeHelper(queue);
newNode.right = deserializeHelper(queue);
return newNode;
}
// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
Queue<String> queue = new LinkedList<>();
queue.addAll(Arrays.asList(data.split(",")));
return deserializeHelper(queue);
}
}
//Codec object will be instantiated and called as such:
//Codec codec = new Codec();
//codec.deserialize(codec.serialize(root));
于 2019-07-16T19:52:17.373 回答