Wednesday, November 2, 2016

LeetCode 449 - Serialize and Deserialize BST


TODO - LeetCode 450 - Delete Node in a BST
https://leetcode.com/problems/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.
https://discuss.leetcode.com/topic/66832/java-o-n-recursive-dfs-without-null-changed-from-serialize-and-deserialize-bt
Thanks to this post, I realize that I can make use of lower and upper bound.
public String serialize(TreeNode root) { // preorder
    StringBuilder sb = new StringBuilder();
    serializedfs(root, sb);
    return sb.toString();
}

private void serializedfs(TreeNode root, StringBuilder sb){
    if(root == null) return; // no "null "
    sb.append(root.val).append(" ");
    serializedfs(root.left, sb);
    serializedfs(root.right, sb);
}

// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
    if(data.length() == 0) return null;
    String[] list = data.split(" ");
    TreeNode dummy = new TreeNode(0);
    deserializedfs(list, 0, dummy, true, Integer.MIN_VALUE, Integer.MAX_VALUE);
    return dummy.left;
}

private int deserializedfs(String[] list, int pos, TreeNode par, boolean isleft, 
                                                    int lower, int upper){
    if(pos >= list.length) return pos;

    int val = Integer.valueOf(list[pos]);
    if(val < lower || val > upper) return pos-1; // have not used this pos, so minus one
    TreeNode cur = new TreeNode(val);
    
    if(isleft) par.left = cur;
    else       par.right = cur;

    pos = deserializedfs(list, ++pos, cur, true, lower, val);
    pos = deserializedfs(list, ++pos, cur, false, val, upper);
    return pos;
}
Using the boundary is really a smart approach. I only rewrote your deserialize part. Here, I will return TreeNode instead of int.
The trick here is to use an array, which stores our current position. I use an int[] array to do so. But it's your choice to pass an int[] parameter or use it as a global variable. :)
    private String splitter = ",";

    // 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 root, StringBuilder sb) {
        if (root == null) return;
        sb.append(root.val).append(splitter);
        buildString(root.left, sb);
        buildString(root.right, sb);
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data.length() == 0) return null;
        int[] pos = new int[1];
        pos[0] = 0;
        return buildTree(data.split(splitter), pos, Integer.MIN_VALUE, Integer.MAX_VALUE);
    }
    
    private TreeNode buildTree(String[] nodes, int[] pos, int min, int max) {
        if (pos[0] == nodes.length) return null;
        
        int val = Integer.valueOf(nodes[pos[0]]);
        if (val < min || val > max) return null; // Go back if we are over the boundary
        TreeNode cur = new TreeNode(val);
        
        pos[0]++; // update current position
        cur.left = buildTree(nodes, pos, min, val);
        cur.right = buildTree(nodes, pos, val, max);
        return cur;
    }
http://www.cnblogs.com/grandyang/p/4913869.html
先序遍历的递归解法,非常的简单易懂,我们需要接入输入和输出字符串流istringstream和ostringstream,对于序列化,我们从根节点开始,如果节点存在,则将值存入输出字符串流,然后分别对其左右子节点递归调用序列化函数即可。对于去序列化,我们先读入第一个字符,以此生成一个根节点,然后再对根节点的左右子节点递归调用去序列化函数即可
    string serialize(TreeNode* root) {
        ostringstream out;
        serialize(root, out);
        return out.str();
    }
    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        istringstream in(data);
        return deserialize(in);
    }
private:
    void serialize(TreeNode *root, ostringstream &out) {
        if (root) {
            out << root->val << ' ';
            serialize(root->left, out);
            serialize(root->right, out);
        } else {
            out << "# ";
        }
    }
    TreeNode* deserialize(istringstream &in) {
        string val;
        in >> val;
        if (val == "#") return nullptr;
        TreeNode *root = new TreeNode(stoi(val));
        root->left = deserialize(in);
        root->right = deserialize(in);
        return root;
    }

http://bookshadow.com/weblog/2016/11/01/leetcode-serialize-and-deserialize-bst/
先序遍历(Preorder Traversal)
根据二叉搜索树(BST)的性质,左孩子 < 根节点 < 右孩子,因此可以通过先序遍历的结果唯一确定一棵原始二叉树。
序列化(Serialization):
先序遍历原始二叉树,输出逗号分隔值字符串。
反序列化(Deserialization):
利用栈(Stack)数据结构,节点栈nstack保存重建二叉树过程中的节点,最大值栈rstack保存当前节点的右孩子允许的最大值。

遍历序列化串,记当前数值为val,新增树节点node = TreeNode(val);记nstack的栈顶元素为ntop(nstack[-1])

若val < ntop,则val为ntop的左孩子,令ntop.left = node,并将node压入nstack

否则,val应为右孩子,但其父节点并不一定为ntop:

    记rstack的栈顶元素为rtop,当val > rtop时,执行循环:
    
        重复弹出nstack,直到ntop不是右孩子为止(nstack[-1] > nstack[-2]条件不成立)
        
        再次弹出nstack, 并弹出rstack

    上述过程执行完毕后,令ntop.right = node,并将node压入nstack
# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Codec: def serialize(self, root): """Encodes a tree to a single string. :type root: TreeNode :rtype: str """ if not root: return [] left = self.serialize(root.left) right = self.serialize(root.right) ans = str(root.val) if left: ans += ',' + left if right: ans += ',' + right return ans def deserialize(self, data): """Decodes your encoded data to tree. :type data: str :rtype: TreeNode """ if not data: return None nstack, rstack = [], [0x7FFFFFFF] for val in map(int, data.split(',')): node = TreeNode(val) if nstack: if val < nstack[-1].val: nstack[-1].left = node rstack.append(nstack[-1].val) else: while val > rstack[-1]: while nstack[-1].val > nstack[-2].val: nstack.pop() rstack.pop() nstack.pop() nstack[-1].right = node nstack.append(node) return nstack[0] # Your Codec object will be instantiated and called as such: # codec = Codec() # codec.deserialize(codec.serialize(root))
X. BFS, Queue
http://www.cnblogs.com/charlesblc/p/6019644.html
class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}

class Codec {
    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        if (root == null) {
            return "";
        }

        // LinkedList实现了Queue接口, ArrayList没有实现
        Queue<TreeNode> qe= new LinkedList<>();
        qe.offer(root);
        sb.append(root.val+",");
        while (!qe.isEmpty()) {
            TreeNode tn = qe.poll();
            if (tn.left != null) {
                sb.append(tn.left.val+",");
                qe.offer(tn.left);
            }
            else {
                sb.append(",");
            }
            if (tn.right != null) {
                sb.append(tn.right.val+",");
                qe.offer(tn.right);
            }
            else {
                sb.append(",");
            }
        }
        return sb.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data.equals("")) {
            return null;
        }

        String[] strs = data.split(",");
        Queue<TreeNode> qe = new LinkedList<>();

        if (strs.length < 1 || strs[0].equals("")) {
            return null;
        }

        TreeNode root = new TreeNode(Integer.valueOf(strs[0]));
        qe.offer(root);
        int i = 1;
        while (!qe.isEmpty()) {
            TreeNode tn = qe.poll();

            if (strs.length > i && !strs[i].equals("")) {
                TreeNode left = new TreeNode(Integer.valueOf(strs[i]));
                tn.left = left;
                qe.offer(left);
            }
            i++;
            if (strs.length > i && !strs[i].equals("")) {
                TreeNode right = new TreeNode(Integer.valueOf(strs[i]));
                tn.right = right;
                qe.offer(right);
            }
            i++;
        }
        return root;
    }
}

另一种方法是层序遍历的非递归解法,这种方法略微复杂一些,我们需要借助queue来做,本质是BFS算法,也不是很难理解,就是BFS算法的常规套路稍作修改即可
    string serialize(TreeNode* root) {
        ostringstream out;
        queue<TreeNode*> q;
        if (root) q.push(root);
        while (!q.empty()) {
            TreeNode *t = q.front(); q.pop();
            if (t) {
                out << t->val << ' ';
                q.push(t->left);
                q.push(t->right);
            } else {
                out << "# ";
            }
        }
        return out.str();
    }
    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        if (data.empty()) return nullptr;
        istringstream in(data);
        queue<TreeNode*> q;
        string val;
        in >> val;
        TreeNode *res = new TreeNode(stoi(val)), *cur = res;
        q.push(cur);
        while (!q.empty()) {
            TreeNode *t = q.front(); q.pop();
            if (!(in >> val)) break;
            if (val != "#") {
                cur = new TreeNode(stoi(val));
                q.push(cur);
                t->left = cur;
            }
            if (!(in >> val)) break;
            if (val != "#") {
                cur = new TreeNode(stoi(val));
                q.push(cur);
                t->right = cur;
            }
        }
        return res;
    }

https://discuss.leetcode.com/topic/66760/java-bfs-solution-easy-to-understand-serialize-deserialize-runtime-o-n
Each tree node can be represented by "val/num/", where val is the value of the node and num indicate his children situation (num == 3 meaning having two children, num == 2 meaning having only left child, num == 1 meaning having only right child, num == 0 meaning having no child).
The time complexity for both serialize and deserialize is O(n), where n is the number of nodes in BST. The trade-off here is that I use an extra char "num" as in val/num/.
    // Encodes a tree to a single string.
 public String serialize(TreeNode root) {
        if (root == null) return "";
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        StringBuilder sb = new StringBuilder();
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            int num = 0;
            if (node.left != null && node.right != null) {
                // 3 indicates having both left and right child
                num = 3;
                queue.offer(node.left);
                queue.offer(node.right);
            } else if (node.left != null) {
                // 2 indicates having left child
                num = 2;
                queue.offer(node.left);
            } else if (node.right != null) {
                // 1 indicates having right child
                num = 1;
                queue.offer(node.right);
            } // 0 indicates having no child
            
            sb.append(node.val).append("/").append(num).append("/");
        }
        return sb.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data == null || data.length() < 4) return null;
        char[] text = data.toCharArray();
        int i = 0;
        TreeNode root = new TreeNode(-1);
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            // Set node value
            i = readNode(text, i, node);
            // Read node's child number
            int num = 0;
            while (text[i] != '/') {
                num = num*10 + Character.getNumericValue(text[i]);
                i++;
            }
            i++;
            if (num == 3) {
                addLeftNode(node, queue);
                addRightNode(node, queue);
            } else if (num == 2) {
                addLeftNode(node, queue);
            } else if (num == 1) {
                addRightNode(node, queue);
            }
        }
        return root;
    }
    
    private int readNode(char[] text, int i, TreeNode node) {
        int val = 0;
        while (text[i] != '/') {
            val = val*10 + Character.getNumericValue(text[i]);
            i++;
        }
        node.val = val;
        return i+1;
    }
    
    private void addLeftNode(TreeNode parent, Queue<TreeNode> queue) {
        TreeNode node = new TreeNode(-1);
        parent.left = node;
        queue.offer(node);
    }
    
    private void addRightNode(TreeNode parent, Queue<TreeNode> queue) {
        TreeNode node = new TreeNode(-1);
        parent.right = node;
        queue.offer(node);
    }
}

X. http://blog.csdn.net/mcf171/article/details/54381539
这个题目可以结合BTS的特点,否则需要保存2n长度的树长度,才知道每一个位置在哪


X. Use extra null nodes
http://www.jianshu.com/p/aa71751524dd
private static final String spliter = " ";
    private static final String NN = "#";

    // 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;
        }
    }
}


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