通過分析解決前面的面試題 6。我們知道可以從前序遍歷和中序遍歷構(gòu)造出一棵二叉樹。受此啟發(fā),我們可以先把一棵二叉樹序列化成一個(gè)前序遍歷序列和一個(gè)中序序列,然后再反序列化時(shí)通過這兩個(gè)序列重構(gòu)出原二叉樹。
這個(gè)思路有兩個(gè)缺點(diǎn)。一個(gè)缺點(diǎn)是該方法要求二叉樹中不能用有數(shù)值重復(fù)的結(jié)點(diǎn)。另外只有當(dāng)兩個(gè)序列中所有數(shù)據(jù)都讀出后才能開始反序列化。如果兩個(gè)遍歷序列的數(shù)據(jù)是從一個(gè)流里讀出來的,那就可能需要等較長的時(shí)間。
實(shí)際上如果二叉樹的序列化是從根結(jié)點(diǎn)開始的話,那么相應(yīng)的反序列化在根結(jié)點(diǎn)的數(shù)值讀出來的時(shí)候就可以開始了。因此我們可以根據(jù)前序遍歷的順序來序列化二叉樹,因?yàn)榍靶虮闅v是從根結(jié)點(diǎn)開始的。當(dāng)在遍歷二叉樹碰到 NULL 指針時(shí),這些 NULL 指針序列化成一個(gè)特殊的字符(比如‘$’)。另外,結(jié)點(diǎn)的數(shù)值之間要用一個(gè)特殊字符(比如’,’)隔開。
private static class BinaryTreeNode {
private int val;
private BinaryTreeNode left;
private BinaryTreeNode right;
public BinaryTreeNode() {
}
public BinaryTreeNode(int val) {
this.val = val;
}
@Override
public String toString() {
return val + "";
}
}
import java.util.LinkedList;
import java.util.List;
public class Test62 {
private static class BinaryTreeNode {
private int val;
private BinaryTreeNode left;
private BinaryTreeNode right;
public BinaryTreeNode() {
}
public BinaryTreeNode(int val) {
this.val = val;
}
@Override
public String toString() {
return val + "";
}
}
public static void serialize(BinaryTreeNode root, List<Integer> result) {
List<BinaryTreeNode> list = new LinkedList<>();
list.add(root);
BinaryTreeNode node;
while (list.size() > 0) {
node = list.remove(0);
if (node == null) {
result.add(null);
}else {
result.add(node.val);
list.add(node.left);
list.add(node.right);
}
}
}
public static BinaryTreeNode deserialize(List<Integer> result, int idx) {
if (result.size() < 1 || idx < 0 || result.size() <= idx || result.get(idx) == null) {
return null;
}
BinaryTreeNode root = new BinaryTreeNode(result.get(idx));
root.left = deserialize(result, idx * 2 + 1);
root.right = deserialize(result, idx * 2 + 2);
return root;
}
public static void main(String[] args) {
test01();
}
private static void test01() {
BinaryTreeNode n1 = new BinaryTreeNode(1);
BinaryTreeNode n2 = new BinaryTreeNode(2);
BinaryTreeNode n3 = new BinaryTreeNode(3);
BinaryTreeNode n4 = new BinaryTreeNode(4);
BinaryTreeNode n5 = new BinaryTreeNode(5);
BinaryTreeNode n6 = new BinaryTreeNode(6);
BinaryTreeNode n7 = new BinaryTreeNode(7);
BinaryTreeNode n8 = new BinaryTreeNode(8);
BinaryTreeNode n9 = new BinaryTreeNode(9);
n1.left = n2;
n1.right = n3;
n2.left = n4;
n2.right = n5;
n3.left = n6;
n3.right = n7;
n4.left = n8;
n4.right = n9;
List<Integer> result = new LinkedList<>();
serialize(n1, result);
System.out.println(result);
System.out.println();
BinaryTreeNode root = deserialize(result, 0) ;
print(root);
}
private static void print(BinaryTreeNode root) {
if (root != null) {
print(root.left);
System.out.printf("%-3d", root.val);
print(root.right);
}
}
}
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