序列化二叉树

2021-04-14

剑指 Offer 37. 序列化二叉树

难度困难143

请实现两个函数，分别用来序列化和反序列化二叉树。

示例:

你可以将以下二叉树：

    1
   / \
  2   3
     / \
    4   5

序列化为 "[1,2,3,null,null,4,5]"

注意：本题与主站 297 题相同：https://leetcode-cn.com/problems/serialize-and-deserialize-binary-tree/

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/**
 * 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) {
        return traverse(root);
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        return builde(data);
    }

    private TreeNode builde(String data) {
        if(data.equals("[]"))return null;
        //System.out.println(data);
        data = data.substring(1, data.length() - 1);
        String[] arr = data.split(",");
        Queue<TreeNode> queue = new LinkedList<>();
        if (arr[0].equals("null")) return null;
        TreeNode root = new TreeNode(Integer.parseInt(arr[0]));
        queue.offer(root);
        int index = 1;

        while (index < arr.length) {

            TreeNode node = queue.poll();
            //System.out.println(node.val);
            if (node == null) continue;
        
            String l ="null";
            String r ="null";
            if(index<arr.length){
                l= arr[index++];
            } 
            if(index<arr.length){
                r= arr[index++];
            } 

            if (l.equals("null")) {
                node.left = null;
            } else {
                int lv = Integer.parseInt(l);
                node.left = new TreeNode(lv);
            }
            queue.offer(node.left);
            if (r.equals("null")) {
                node.right = null;
            } else {
                int rv = Integer.parseInt(r);
                node.right = new TreeNode(rv);
            }
            queue.offer(node.right);
        }
        return root;
    }

    private String traverse(TreeNode root) {
        if (root == null) return "[]";
        Queue<TreeNode> queue = new LinkedList<>();
        StringBuilder bd = new StringBuilder();
        bd.append("[");
        queue.offer(root);

        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            if (node == null){
                bd.append("null,");
                continue;
            }
            bd.append(node.val);
            bd.append(",");

            queue.offer(node.left);
            queue.offer(node.right);
        }
        int index = bd.length()-1;

        while(!Character.isDigit(bd.charAt(index--))){
            bd.deleteCharAt(bd.length() - 1);
        }
        bd.append("]");
        return bd.toString();
    }
}

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

public class Codec {
    public String serialize(TreeNode root) {
        if(root == null) return "[]";
        StringBuilder res = new StringBuilder("[");
        Queue<TreeNode> queue = new LinkedList<>() {{ add(root); }};
        while(!queue.isEmpty()) {
            TreeNode node = queue.poll();
            if(node != null) {
                res.append(node.val + ",");
                queue.add(node.left);
                queue.add(node.right);
            }
            else res.append("null,");
        }
        res.deleteCharAt(res.length() - 1);
        res.append("]");
        return res.toString();
    }

    public TreeNode deserialize(String data) {
        if(data.equals("[]")) return null;
        String[] vals = data.substring(1, data.length() - 1).split(",");
        TreeNode root = new TreeNode(Integer.parseInt(vals[0]));
        Queue<TreeNode> queue = new LinkedList<>() {{ add(root); }};
        int i = 1;
        while(!queue.isEmpty()) {
            TreeNode node = queue.poll();
            if(!vals[i].equals("null")) {
                node.left = new TreeNode(Integer.parseInt(vals[i]));
                queue.add(node.left);
            }
            i++;
            if(!vals[i].equals("null")) {
                node.right = new TreeNode(Integer.parseInt(vals[i]));
                queue.add(node.right);
            }
            i++;
        }
        return root;
    }
}

作者：jyd
链接：https://leetcode-cn.com/problems/xu-lie-hua-er-cha-shu-lcof/solution/mian-shi-ti-37-xu-lie-hua-er-cha-shu-ceng-xu-bian-/
来源：力扣（LeetCode）
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