題目描述

請實現兩個函式，分別用來序列化和反序列化二叉樹

解題思路

使用前序遍歷，將遇到的結點新增到字串中，遇到null則將一個#新增要序列化字串中。反序列化時，每次讀取根結點，然後讀取其左結點，遇到#（null）時，返回上層。

實現

/*樹結點定義*/
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
/*實現*/
 

public class Solution {
    String Serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        serialize(root, sb);
        return sb.toString();
    }

    private void serialize(TreeNode root, StringBuilder sb) {
        if (root == null) {
            sb.append("#,");
            return
 
;
        }

        sb.append(root.val + ",");
        serialize(root.left, sb);
        serialize(root.right, sb);
    }

    private class Result{
        TreeNode node;
        int pos;

        Result(TreeNode node, int pos){
            this.node = node;
            this.pos = pos;
        }
    }

    TreeNode Deserialize(String str) {
        if
 
 (str == null || str.length() <= 0) return null;
        String[] strs = str.split(",");
        Result re = deserialize(strs, 0);
        return re.node;
    }

    private Result deserialize(String[] str, int i) {
        TreeNode root = null;
        if (i < str.length - 1){
            if ("#".equals(str[i])) return new Result(null, i+1);
            root = new TreeNode(Integer.parseInt(str[i]));
            Result l = deserialize(str, i + 1);
            root.left = l.node;
            Result r = deserialize(str, l.pos);
            root.right = r.node;
            return new Result(root, r.pos);
        }
        return new Result(root, i+1);
    }
}