[LeetCode]297. Serialize and Deserialize Binary Tree

最新推荐文章于 2024-06-04 10:47:19 发布

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最新推荐文章于 2024-06-04 10:47:19 发布

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本文介绍了一种实现二叉树序列化和反序列化的算法。通过队列遍历的方式，将二叉树节点的信息编码为字符串进行序列化，并能从该字符串重构出原始的二叉树结构。

https://leetcode.com/problems/serialize-and-deserialize-binary-tree/

https://leetcode.com/problems/serialize-and-deserialize-bst/

序列化和反序列化二叉树

public class Codec {

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        Queue<TreeNode> queue = new LinkedList();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                if (node == null) {
                    sb.append(",null");
                } else {
                    sb.append("," + node.val);
                    queue.offer(node.left);
                    queue.offer(node.right);
                }
            }
        }
        sb.deleteCharAt(0);
        return sb.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        String[] arr = data.split(",");
        if (arr[0].equals("null")) {
            return null;
        }
        TreeNode root = new TreeNode(Integer.parseInt(arr[0]));
        Queue<TreeNode> queue = new LinkedList();
        queue.offer(root);
        int index = 1;
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                if (node == null) {
                    continue;
                }
                TreeNode left = arr[index].equals("null") ? null : new TreeNode(Integer.parseInt(arr[index]));
                index++;
                TreeNode right = arr[index].equals("null") ? null : new TreeNode(Integer.parseInt(arr[index]));
                index++;
                node.left = left;
                node.right = right;
                queue.offer(left);
                queue.offer(right);
            }
        }
        return root;
    }
}