LeetCode 297. Serialize and Deserialize Binary Tre

cc17 发布于2019-08-16 13:28 / 3193人阅读

摘要：题目大意将二叉树序列化，返回序列化的，和反序列化还原。解题思路技巧在于将记录为便于将来判断。的思想将每一层记录下来，反序列化时也按照层级遍历的方法依次设置为上一个里面的元素的左孩子和右孩子。变种，可以要求输出一个，而不是

LeetCode 297. Serialize and Deserialize Binary Tree

题目大意: 将二叉树序列化，返回序列化的String，和反序列化还原。

解题思路：
技巧在于将null记录为#便于将来判断。有两种解法。

Level Order Traversal - BFS的思想将每一层记录下来，反序列化时也按照层级遍历的方法依次设置为上一个queue里面的元素的左孩子和右孩子。

更好的方法为preorder traversal，是可以handle变种题目的解法，利用preorder是root—>left->right的顺序，用一个deque来不断把头部的元素poll出，递归调用函数构建还原二叉树。

</>复制代码 
    //Solution 1: using Level order traversal
    public static String serialize(TreeNode root) {
        if (root == null ) {
            return "";
        }
        StringBuilder sb = new StringBuilder();
        Queue q = new LinkedList();
        q.offer(root);
        while(!q.isEmpty()) {
            TreeNode curr = q.poll();
            if (curr == null) {
                sb.append("#,");
            } else {
                sb.append(curr.val + ",");
                q.offer(curr.left);
                q.offer(curr.right);
            }
        }
        sb.deleteCharAt(sb.length() - 1);
        return sb.toString();
    }
    /**
     *      1
     *     / 
     *    2   3
     *   /   / 
     *  4   2   4
     *  /
     * 4
     * [1,2,3,4,#,2,4,4]
     * @param input
     * @return
     */
    public static TreeNode deSerialize(String input) {
        if (input == null || input.length() == 0) {
            return null;
        }
        String[] strs = input.split(",");
        TreeNode root = new TreeNode(Integer.valueOf(strs[0]));
        Queue q = new LinkedList();
        q.offer(root);
        for(int i = 1; i < strs.length; i++){
            TreeNode curr = q.poll();
            if (curr == null) continue;
            if (!strs[i].equals("#")) {
                curr.left = new TreeNode(Integer.valueOf(strs[i]));
                q.offer(curr.left);
            }
            i++;
            if (i < strs.length && !strs[i].equals("#")) {
                curr.right = new TreeNode(Integer.valueOf(strs[i]));
                q.offer(curr.right);
            }
        }
        return root;
    }
    //Solution II: using 2 preorder traversal
    private static final String DELIMITER = ",";
    private static final String NN = "#";
    // Encodes a tree to a single string.
    public static String serialize2(TreeNode root) {
        //using preorder traversal
        StringBuilder res = new StringBuilder();
        serializeHelper(root, res);
        res.deleteCharAt(res.length() - 1);
        return res.toString();
    }
    private static void serializeHelper(TreeNode root, StringBuilder res) {
        if (root == null) {
            res.append(NN).append(DELIMITER);
            return;
        }
        res.append(root.val).append(DELIMITER);
        serializeHelper(root.left, res);
        serializeHelper(root.right, res);
    }
    // Decodes your encoded data to tree.
    public static TreeNode deserialize2(String data) {
        Deque deque = new LinkedList<>();
        deque.addAll(Arrays.asList(data.split(DELIMITER)));
        return deserializeHelper(deque);
    }
    private static TreeNode deserializeHelper(Deque deque) {
        String now = deque.pollFirst();
        if (now.equals(NN)) {
            return null;
        }
        TreeNode node = new TreeNode(Integer.parseInt(now));
        node.left = deserializeHelper(deque);
        node.right = deserializeHelper(deque);
        return node;
    }
    public static void main(String[] args) {
        //solution 1 level order
        TreeNode root1 = deSerialize("1,2,3,4,#,5,6,7");
        String result1 = serialize(root1);
        System.out.println(result1);
        //solution 2 preorder - 变种，可以要求输出一个LinkedList，而不是String
        TreeNode root2 = deserialize2("3,4,1,#,#,2,#,#,5,#,6,#,#");
        String result2 = serialize2(root2);
        System.out.println(result2);
    }

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