🌕🌕297. 二叉树的序列化与反序列化

难度: 🌕🌕

问题描述

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解法 1 - 层序

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Codec {
    // 思路：
    // 层序遍历 - 序列化
    // 空节点也入队

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        if(root == null) {
            return sb.toString();
        }
        LinkedList<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()) {
            int len = queue.size();
            for(int i = 0; i < len; i ++) {
                TreeNode cur = queue.poll();
                if(cur == null) {
                    sb.append('#').append(",");
                } else {
                    sb.append(cur.val).append(",");
                    queue.offer(cur.left);
                    queue.offer(cur.right);
                }
            }
        }
        sb.deleteCharAt(sb.length() - 1);
        return sb.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if(data.length() == 0) {
            return null;
        }
        String[] array = data.split(",");
        TreeNode root = new TreeNode(Integer.valueOf(array[0]));
        // 仍然借助 辅助队列
        LinkedList<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int index = 1; // 除去根节点的 下标 0，从 1 开始
        while(!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            String left = array[index];
            index ++;
            String right = array[index];
            index ++;
            if(!left.equals("#")) {
                cur.left = new TreeNode(Integer.valueOf(left));
                queue.offer(cur.left);
            }
            if(!right.equals("#")) {
                cur.right = new TreeNode(Integer.valueOf(right));
                queue.offer(cur.right);
            }
        }
        return root;
    }
}

// Your Codec object will be instantiated and called as such:
// Codec ser = new Codec();
// Codec deser = new Codec();
// TreeNode ans = deser.deserialize(ser.serialize(root));

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输出 1

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解法 2 - 递归

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Codec {
    // 思路：
    // 先序遍历 - 空节点也入结果集
    int index = 0;

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        if(root == null) {
            return "";
        }
        StringBuilder sb = new StringBuilder();
        mySol1(root, sb);
        sb.deleteCharAt(sb.length() - 1);
        return sb.toString();
    }

    private void mySol1(TreeNode root, StringBuilder sb) {
        // 递归终止条件
        if(root == null) {
            sb.append("#").append(",");
            return;
        }
        sb.append(root.val).append(",");
        mySol1(root.left, sb);
        mySol1(root.right, sb);
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if(data.length() == 0) {
            return null;
        }
        String[] array = data.split(",");
        index = 0;
        return mySol2(array);
    }

    private TreeNode mySol2(String[] array) {
        String cur = array[index];
        if(cur.equals("#")) {
            index ++;
            return null;
        }
        int val = Integer.valueOf(cur);
        TreeNode res = new TreeNode(val);
        index ++;
        res.left = mySol2(array);
        res.right = mySol2(array);
        return res;
    }
}

// Your Codec object will be instantiated and called as such:
// Codec ser = new Codec();
// Codec deser = new Codec();
// TreeNode ans = deser.deserialize(ser.serialize(root));

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输出 2