LeetCode – Serialize and Deserialize Binary Tree (Java)

 

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.

Java Solution 1 - Level Order Traveral

// Encodes a tree to a single string.
public String serialize(TreeNode root) {
    if(root==null){
        return "";
    }
 
    StringBuilder sb = new StringBuilder();
 
    LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
 
    queue.add(root);
    while(!queue.isEmpty()){
        TreeNode t = queue.poll();
        if(t!=null){
            sb.append(String.valueOf(t.val) + ",");
            queue.add(t.left);
            queue.add(t.right);
        }else{
            sb.append("#,");
        }
    }
 
    sb.deleteCharAt(sb.length()-1);
    System.out.println(sb.toString());
    return sb.toString();
}
 
// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
    if(data==null || data.length()==0)
        return null;
 
    String[] arr = data.split(",");
    TreeNode root = new TreeNode(Integer.parseInt(arr[0]));
 
 
    LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
    queue.add(root);
 
    int i=1;
    while(!queue.isEmpty()){
        TreeNode t = queue.poll();
 
        if(t==null)
            continue;
 
        if(!arr[i].equals("#")){
            t.left = new TreeNode(Integer.parseInt(arr[i]));    
            queue.offer(t.left);
 
        }else{
            t.left = null;
            queue.offer(null);
        }
        i++;
 
        if(!arr[i].equals("#")){
            t.right = new TreeNode(Integer.parseInt(arr[i]));    
            queue.offer(t.right);
 
        }else{
            t.right = null;
            queue.offer(null);
        }
        i++;
    }
 
    return root;
}

Java Solution 2 - Preorder Traversal

// Encodes a tree to a single string.
public String serialize(TreeNode root) {
    if(root==null)
        return null;
 
    Stack<TreeNode> stack = new Stack<TreeNode>();
    stack.push(root);
    StringBuilder sb = new StringBuilder();
 
    while(!stack.isEmpty()){
        TreeNode h = stack.pop();   
        if(h!=null){
            sb.append(h.val+",");
            stack.push(h.right);
            stack.push(h.left);  
        }else{
            sb.append("#,");
        }
    }
 
    return sb.toString().substring(0, sb.length()-1);
}
 
// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
    if(data == null)
        return null;
 
    int[] t = {0};
    String[] arr = data.split(",");
 
    return helper(arr, t);
}
 
public TreeNode helper(String[] arr, int[] t){
    if(arr[t[0]].equals("#")){
        return null;
    }
 
    TreeNode root = new TreeNode(Integer.parseInt(arr[t[0]]));
 
    t[0]=t[0]+1;
    root.left = helper(arr, t);
    t[0]=t[0]+1;
    root.right = helper(arr, t);
 
    return root;
}

Related posts:

  1. LeetCode – Binary Search Tree Iterator (Java)
  2. LeetCode – Binary Tree Right Side View (Java)
  3. LeetCode – Balanced Binary Tree (Java)
  4. LeetCode – Invert Binary Tree (Java)
Category >> Algorithms  
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  1. carlos on 2016-10-30

    I’m confused as to why the second solution works, since java is passed by value how would the T get updated?

  2. Yi Ling on 2016-12-23

    t is an array with one element. When the recursion function changed the t inside. The t outside will be modified at the same time. Since they share the same memory space. In another word, the reference t inside is different from the reference t outside, but they point to the same memory space.

  3. Codingpan on 2016-12-26

    In the while loop, can we remove queue.offer(null) when string in the arr is “#”? so it doesn’t need to check t=null when we poll element from queue. Is there any reason to add null into the queue?


    while(!queue.isEmpty()){
    TreeNode t = queue.poll();

    if(!arr[i].equals("#")){
    t.left = new TreeNode(Integer.parseInt(arr[i]));
    queue.offer(t.left);
    }else{
    t.left = null;
    }
    i++;

    if(!arr[i].equals("#")){
    t.right = new TreeNode(Integer.parseInt(arr[i]));
    queue.offer(t.right);
    }else{
    t.right = null;
    }
    i++;
    }

  4. Shubham Barhate on 2017-3-13

    why have we used an array t of size 1 and increading the value of 0th index, instead of just using a variable and incrementing it..

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