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 5as
"[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.
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.
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