Buttercola: Leetcode: Serialize and Deserialize Binary Tree

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

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.

Update on 1/28/16:
BFS by removing the trailing "null" node strings.
One advantage of using BFS is we could safely removing the trailing last-level null nodes represented by the "#," string. This can save about half space (The # nodes in last-level of a binary tree is around half of the total number of trees).

Code (Java):

/**
 * 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) {
        if (root == null) {
            return "";
        }
         
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
         
        StringBuffer sb = new StringBuffer();
         
        while (!queue.isEmpty()) {
            TreeNode curr = queue.poll();
            if (curr != null) {
                sb.append(curr.val + ",");
                queue.offer(curr.left);
                queue.offer(curr.right);
            } else {
                sb.append("#" + ",");
            }
        }
         
        // Remove the trailing #
        String result = sb.toString();
        int j = result.length() - 1;
         
        while (j > 0 && result.charAt(j) == ',' && result.charAt(j) == '#') {
            j -= 2;
        }
         
        result = result.substring(0, j);
        return sb.toString();
    }
 
    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data == null || data.length() == 0) {
            return null;
        }
         
        Queue<TreeNode> queue = new LinkedList<>();
        String[] nodes = data.split(",");
         
        TreeNode root = new TreeNode(Integer.parseInt(nodes[0]));
        queue.offer(root);
        int i = 1;
 
        while (!queue.isEmpty() && i < nodes.length) {
            TreeNode curr = queue.poll();
            if (nodes[i].equals("#")) {
                curr.left = null;
            } else {
                TreeNode leftNode = new TreeNode(Integer.parseInt(nodes[i]));
                curr.left = leftNode;
                queue.offer(leftNode);
            }
             
            i++;
            if (i >= nodes.length) {
                break;
            }
             
            // right node
            if (nodes[i].equals("#")) {
                curr.right = null;
            } else {
                TreeNode rightNode = new TreeNode(Integer.parseInt(nodes[i]));
                curr.right = rightNode;
                queue.offer(rightNode);
            }
             
            i++;
        }
         
        return root;
    }
}
 
// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.deserialize(codec.serialize(root));

DFS solution with using a Deque:
We can also solve the problem by using pre-order traversal of the binary tree. We need to use a deque because each time we need to remove one element from the top. Else we need to maintain a global index pointing to the current element we want to add into the tree.

Code (Java):

/**
 * 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) {
        if (root == null) {
            return "";
        }
         
        StringBuffer sb = new StringBuffer();
        serializeHelper(root, sb);
         
        return sb.toString();
    }
     
    private void serializeHelper(TreeNode root, StringBuffer sb) {
        if (root == null) {
            sb.append("#");
            sb.append(",");
            return;
        }
         
        sb.append(root.val + ",");
        serializeHelper(root.left, sb);
        serializeHelper(root.right, sb);
    }
 
    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data == null || data.length() == 0) {
            return null;
        }
         
        String[] strs = data.split(",");
        Deque<String> deque = new LinkedList<>(Arrays.asList(strs));
         
        return deserializeHelper(deque);
    }
     
    private TreeNode deserializeHelper(Deque<String> deque) {
        if (deque.isEmpty()) {
            return null;
        }
         
        String node = deque.removeFirst();
        if (node.equals("#")) {
            return null;
        }
         
        TreeNode root = new TreeNode(Integer.parseInt(node));
        root.left = deserializeHelper(deque);
        root.right = deserializeHelper(deque);
         
        return root;
    }
}
 
// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.deserialize(codec.serialize(root));

Another DFS without using a Deque, but a global index.

public class Solution2 {
    public String serialdfs(TreeNode root) {
        if (root == null) {
            return "";
        }
         
        StringBuilder sb = new StringBuilder();
         
        serialdfsHelper(root, sb);
         
        return sb.toString();
    }
     
    private void serialdfsHelper(TreeNode root, StringBuilder sb) {
        if (root == null) {
            sb.append("#");
            return;
        }
         
        sb.append(root.val);
         
        serialdfsHelper(root.left, sb);
        serialdfsHelper(root.right, sb);
    }
     
    public int i = 0;
    public TreeNode deserialdfs(String s) {
        if (s == null || s.length() == 0) {
            return null;
        }
         
        return deserialdfsHelper(s);
    }
     
    private TreeNode deserialdfsHelper(String s) {
        if (i >= s.length() || s.charAt(i) == '#') {
            return null;
        }
         
        TreeNode root = new TreeNode(s.charAt(i) - '0');
        i++;
        root.left = deserialdfsHelper(s);
        i++;
        root.right = deserialdfsHelper(s);
         
        return root;
    }
     
    public static void main(String[] args) {
      Solution2 sol = new Solution2();
 
      TreeNode root = new TreeNode(1);
      root.left = new TreeNode(2);
      root.right = new TreeNode(3);
      root.left.left = new TreeNode(4);
      root.left.right = new TreeNode(5);
 
 
      String result = sol.serialdfs(root);
       
      TreeNode root2 = sol.deserialdfs(result);
 
      String result2 = sol.serialdfs(root2);
 
      System.out.println(result2);
    }
}

Compare between DFS and BFS:
I would like more BFS because BFS sometimes can save more space. For example,
1
/ \
2 3
/ \ / \
4 # # #
/ \
# #

By BFS: the serialized tree is 1,2,3,4
By DFS, the serialized tree is 1,2,4,#,#,#,3, which needs more space to store.

Buttercola

Tuesday, October 27, 2015

Leetcode: Serialize and Deserialize Binary Tree

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