🌕🌕 剑指 Offer 37. 序列化二叉树

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 - 空节点也入队列

    // 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<>();
        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(',');
                    continue;
                }
                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) {
        int len = data.length();
        if(len == 0) {
            return null;
        }
        String[] arr = data.split(",");
        TreeNode root = new TreeNode(Integer.valueOf(arr[0]));
        int index = 1;
        LinkedList<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            String l = arr[index];
            if(l.equals("#")) {
                index ++;
            } else {
                TreeNode left = new TreeNode(Integer.valueOf(l));
                cur.left = left;
                queue.offer(left);
                index ++;
            }
            String r = arr[index];
            if(r.equals("#")) {
                index ++;
            } else {
                TreeNode right = new TreeNode(Integer.valueOf(r));
                cur.right = right;
                queue.offer(right);
                index ++;
            }
        }
        return root;
    }
}

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

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输出 1

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解法 2 - 先序

/**
 * 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：
    // 先序遍历 - 空节点也入队
    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;
        }
        // root != null
        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.equals("")) {
            return null;
        }
        String[] arr = data.split(",");
        int len = arr.length;
        return mySol2(arr, len);
    }
    private TreeNode mySol2(String[] arr, int len) {
        // 递归终止条件
        if(index == len) {
            return null;
        }
        String cur = arr[index]; // 根节点
        if(cur.equals("#")) {
            // 说明是个 空节点
            index ++;
            return null;
        }
        // 非空节点
        TreeNode res = new TreeNode(Integer.valueOf(cur));
        index ++;
        res.left = mySol2(arr, len);
        res.right = mySol2(arr, len);
        return res;
    }
}

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

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输出 2