leetcode 449. Serialize and Deserialize BST

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 search tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary search tree can be serialized to a string and this string can be deserialized to the original tree structure.

The encoded string should be as compact as possible.

Note: Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.

然后给出的方法模板是这样的：

/** * 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) {             }    // Decodes your encoded data to tree.    public TreeNode deserialize(String data) {                }}// Your Codec object will be instantiated and called as such:// Codec codec = new Codec();// codec.deserialize(codec.serialize(root));

如何用字符串来表示一棵树呢？我就想到了用 [ val , ( leftTree , rightTree ) ] 的形式。如果没有左右子树，那么为 [ val  ] 。如果没有右子树，那么为 [ val , ( leftTree ) ] 。

举个例子。下面这张图：

对应的string就是 [1([2(,[4])],[3([5])])]

下面是我的解法：

//Encodes a tree to a single string.public String serialize(TreeNode root) {     if(root==null){return "";}if(root.left==null&&root.right==null){return "["+root.val+"]";}else if(root.right==null){return "["+root.val+"("+serialize(root.left)+")]";}else{return "["+root.val+"("+serialize(root.left)+","+serialize(root.right)+")]";}}// Decodes your encoded data to tree.public TreeNode deserialize(String data) {        if(data==null||data.equals("")){return null;}if(!data.contains("(")){data=data.substring(1,data.length()-1);//把两侧的[ ]剥掉int val=Integer.parseInt(data);return new TreeNode(val);}int index=data.indexOf('(');String valString=data.substring(1,index);    String subRootString=data.substring(index+1,data.length()-2);//去掉(和)] int val=Integer.parseInt(valString); TreeNode root=new TreeNode(val); int i=0;//i是要得出当前括号所对应逗号的indexint countLeft=0;while(i<subRootString.length()){if(subRootString.charAt(i)=='['){countLeft++;}else if(subRootString.charAt(i)==']'){countLeft--;}else if(countLeft==0&&subRootString.charAt(i)==','){break;}i++;}    if(i==subRootString.length()){//说明当前括号没有对应逗号，即当前只有左结点    root.left=deserialize(subRootString);    }    else{    String left=subRootString.substring(0, i);    String right=subRootString.substring(i+1);    root.left=deserialize(left);    root.right=deserialize(right);    }return root;   }

大神是用先根遍历+队列来做的，思路如下：

BST的先根遍历是先输出根结点，再输出左、右结点。

root left1 left2 leftX right1 rightX

BST的特征是它的左子树的值都小于它，它的右子树的值都大于它。那么我们注意先根遍历的值：

rootValue (<rootValue) (<rootValue) (<rootValue) |separate line| (>rootValue) (>rootValue)

那么，在分割线 | 前面的值都比当前根的值小，在分割线| 后面的值都比当前根的值大。我们根据这个特性来构造左右子树。我们使用队列来获得根节点、左右子树。

public class Codec {    private static final String SEP = ",";    private static final String NULL = "null";    // Encodes a tree to a single string.    public String serialize(TreeNode root) {        StringBuilder sb = new StringBuilder();        if (root == null) return NULL;        //traverse it recursively if you want to, I am doing it iteratively here        Stack<TreeNode> st = new Stack<>();        st.push(root);        while (!st.empty()) {            root = st.pop();            sb.append(root.val).append(SEP);            if (root.right != null) st.push(root.right);            if (root.left != null) st.push(root.left);        }        return sb.toString();    }    // Decodes your encoded data to tree.    // pre-order traversal    public TreeNode deserialize(String data) {        if (data.equals(NULL)) return null;        String[] strs = data.split(SEP);        Queue<Integer> q = new LinkedList<>();        for (String e : strs) {            q.offer(Integer.parseInt(e));        }        return getNode(q);    }        // some notes:    //   5    //  3 6    // 2   7    private TreeNode getNode(Queue<Integer> q) { //q: 5,3,2,6,7        if (q.isEmpty()) return null;        TreeNode root = new TreeNode(q.poll());//root (5)        Queue<Integer> samllerQueue = new LinkedList<>();        while (!q.isEmpty() && q.peek() < root.val) {            samllerQueue.offer(q.poll());        }        //smallerQueue : 3,2   storing elements smaller than 5 (root)        root.left = getNode(samllerQueue);        //q: 6,7   storing elements bigger than 5 (root)        root.right = getNode(q);        return root;    }}