Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.
For example, you may serialize the following tree
1
/ \
2 3
/ \
4 5
as "[1,2,3,null,null,4,5]", just the same as how LeetCode OJ serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.
Note: Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.
Credits:
Special thanks to @Louis1992 for adding this problem and creating all test cases.
【解析1】
其实LeetCode上树的表示方式就挺好，即"[1,2,3,null,null,4,5]"这种形式，我们接下来就实现以下这种序列化。
序列化比较容易，我们做一个层次遍历就好，空的地方用null表示，稍微不同的地方是题目中示例得到的结果是"[1,2,3,null,null,4,5,null,null,null,null,]"，即 4 和 5 的两个空节点我们也存了下来。
饭序列化时，我们根据都好分割得到每个节点。需要注意的是，反序列化时如何寻找父节点与子节点的对应关系，我们知道在数组中，如果满二叉树（或完全二叉树）的父节点下标是 i，那么其左右孩子的下标分别为 2*i+1 和 2*i+2，但是这里并不一定是满二叉树（或完全二叉树），所以这个对应关系需要稍作修改。如下面这个例子：
5
/ \
4 7
/ /
3 2
/ /
-1 9
序列化结果为[5,4,7,3,null,2,null,-1,null,9,null,null,null,null,null,]。
其中，节点 2 的下标是 5，可它的左孩子 9 的下标为 9，并不是 2*i+1=11，原因在于 前面有个 null 节点，这个 null 节点没有左右孩子，所以后面的节点下标都提前了2。所以我们只需要记录每个节点前有多少个 null 节点，就可以找出该节点的孩子在哪里了，其左右孩子分别为 2*(i-num)+1 和 2*(i-num)+2（num为当前节点之前 null 节点的个数）。
【java代码】非递归
/**
* 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) {
StringBuilder sb = new StringBuilder();
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root); while (!queue.isEmpty()) {
TreeNode node = queue.poll();
if (node == null) {
sb.append("null,");
} else {
sb.append(String.valueOf(node.val) + ",");
queue.offer(node.left);
queue.offer(node.right);
}
} return sb.toString();
} // Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
if (data == null || data.isEmpty()) return null; String[] vals = data.split(",");
int[] nums = new int[vals.length]; // 节点i之前null节点的个数
TreeNode[] nodes = new TreeNode[vals.length]; for (int i = 0; i < vals.length; i++) { //计算每个节点前面null节点的数目
if (i > 0) {
nums[i] = nums[i - 1];
}
if (vals[i].equals("null")) {
nodes[i] = null;
nums[i]++;
} else {
nodes[i] = new TreeNode(Integer.parseInt(vals[i]));
}
} for (int i = 0; i < vals.length; i++) { //对节点进行连接操作
if (nodes[i] == null) {
continue;
}
nodes[i].left = nodes[2 * (i - nums[i]) + 1];
nodes[i].right = nodes[2 * (i - nums[i]) + 2];
} return nodes[0];
} }
【解析2】
我们也可以用递归来解决这个问题：The idea is simple: print the tree in pre-order traversal and use "X" to denote null node and split node with ",". We can use a StringBuilder for building the string on the fly. For deserializing, we use a Queue to store the pre-order traversal and since we have "X" as null node, we know exactly how to where to end building subtress.
这个思路用到的是树的前序遍历，因为序列中包含了值为null的节点，因此我们可以很容易地进行反序列化操作。
【java代码】递归
public class Codec {
private static final String spliter = ",";
private static final String NN = "X";
// Encodes a tree to a single string.
public String serialize(TreeNode root) {
StringBuilder sb = new StringBuilder();
buildString(root, sb);
return sb.toString();
}
private void buildString(TreeNode node, StringBuilder sb) {
if (node == null) {
sb.append(NN).append(spliter);
} else {
sb.append(node.val).append(spliter);
buildString(node.left, sb);
buildString(node.right,sb);
}
}
// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
Deque<String> nodes = new LinkedList<>();
nodes.addAll(Arrays.asList(data.split(spliter)));
return buildTree(nodes);
}
private TreeNode buildTree(Deque<String> nodes) {
String val = nodes.remove();
if (val.equals(NN)) return null;
else {
TreeNode node = new TreeNode(Integer.valueOf(val));
node.left = buildTree(nodes);
node.right = buildTree(nodes);
return node;
}
}
}