【剑指offer】面试题37：序列化二叉树

题目

请实现两个函数，分别用来序列化和反序列化二叉树

思路

使用前序遍历来序列化二叉树，当碰到子节点为null的时候，用特殊字符’$’来表示。

ps：也可以使用后序遍历（左->右->根）来做。但是不能使用中序遍历来做（左->根->右），因为使用中序遍历无法确定这棵二叉树。

代码

public class _37_SerializeBinaryTrees {    public static class TreeNode {        public int val = 0;        public TreeNode left = null;        public TreeNode right = null;        public TreeNode(int val) {            this.val = val;        }    }    /**     * 序列化     */    public String Serialize(TreeNode root) {        if(root == null)            return "$";        StringBuilder sb = new StringBuilder();        Serialize(root, sb);        return sb.toString();    }    private void Serialize(TreeNode root, StringBuilder sb) {        if(root == null) {            sb.append("$");            return;        }        sb.append(root.val);        Serialize(root.left, sb.append(","));        Serialize(root.right, sb.append(","));    }    private int index;    /**     * 反序列化     */    public TreeNode Deserialize(String str) {        if(str == null || str.trim().length() == 0)            return null;        String[] nums = str.split(",");        index = 0;        return Deserialize(nums);    }    public TreeNode Deserialize(String[] nums) {        if(nums[index].equals("$")) {            ++index;            return null;        }        TreeNode node = new TreeNode(Integer.valueOf(nums[index++]));        node.left = Deserialize(nums);        node.right = Deserialize(nums);        return node;    }}

测试

public class _37_Test {    public static void main(String[] args) {        test1();        test2();        test3();    }    /**     * 功能测试     * 1.序列化      *     1     *    / \     *   2   3     *  /   / \     * 4   5   6     *      * 2.反序列化     */    private static void test1() {        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);        TreeNode node6 = new TreeNode(6);        node1.left = node2;        node1.right = node3;        node2.left = node4;        node3.left = node5;        node3.right = node6;        _37_SerializeBinaryTrees sbt = new _37_SerializeBinaryTrees();        String s = sbt.Serialize(node1);        // 1,2,4,$,$,$,3,5,$,$,6,$,$        System.out.println(s);        TreeNode newRoot = sbt.Deserialize(s);        String s2 = sbt.Serialize(newRoot);        // 1,2,4,$,$,$,3,5,$,$,6,$,$        System.out.println(s);    }    /**     * 边界测试      * 1.只有左子节点         * 2.只有右子节点     * 3.只有一个节点     */    private static void test2() {        _37_SerializeBinaryTrees sbt = new _37_SerializeBinaryTrees();        TreeNode root1 = sbt.Deserialize("1,2,3,$,$,$,$");        System.out.println(sbt.Serialize(root1));        TreeNode root2 = sbt.Deserialize("1,$,2,$,3,$,$");        System.out.println(sbt.Serialize(root2));        TreeNode root3 = sbt.Deserialize("1,$,$");        System.out.println(sbt.Serialize(root3));    }    /**     * 极端测试     * 1.反序列化时输入空字符串     * 2.序列化时输入null     */    private static void test3() {        _37_SerializeBinaryTrees sbt = new _37_SerializeBinaryTrees();        TreeNode root = sbt.Deserialize("");        System.out.println(sbt.Serialize(root));    }}