剑指offer-37 序列化二叉树

题目：

思路：

自己解答：

这个有错误

public class Solution {
    String Serialize(TreeNode root) {
        if(root == null) return "#!";
        StringBuilder bd = new StringBuilder();
        serializeCore(root, bd);
        return bd.toString();
  }
    private void serializeCore(TreeNode root, StringBuilder bd){
        if(root == null){
            bd.append("#!");
            return;
        }
        bd.append(root.val);
        bd.append("!");
        serializeCore(root.left, bd);
        serializeCore(root.right, bd);
    }

    TreeNode Deserialize(String str) {
       if(str == null || str.length() <= 1) return null;
        TreeNode res = null;
        deserializeCore(res, str, 0);
        return res;
  }
    private int end;
    private void deserializeCore(TreeNode res, String str, int start){
        end = str.indexOf("!", start);
        if(end == -1) return;
        String num = str.substring(start, end);
        if(num.equals("#")) return;
        res = new TreeNode(Integer.parseInt(num));
        res.left = null;
        res.right = null;
        deserializeCore(res.left, str, ++end);
        deserializeCore(res.right, str, ++end);
    }
}

犯的错误：

反序列化时不能正确的链接各个节点

注意：

1. 引用类型赋值为null时，无论里面怎么修改，出来还是null
2. 对于这种从上往下需要链接，可以像下面代码一样利用return返回来链接
3. String,StringBuffer,StringBuilder三者区别https://www.cnblogs.com/weibanggang/p/9455926.html

别人解答：

/*
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

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

    }

}
*/
public class Solution {
    public int index = -1;
    String Serialize(TreeNode root) {
        StringBuffer sb = new StringBuffer();
        if(root == null){
            sb.append("#,");
            return sb.toString();
        }
        sb.append(root.val + ",");
        sb.append(Serialize(root.left));
        sb.append(Serialize(root.right));
        return sb.toString();
  }
    TreeNode Deserialize(String str) {
        index++;
       int len = str.length();
        if(index >= len){
            return null;
        }
        String[] strr = str.split(",");
        TreeNode node = null;
        if(!strr[index].equals("#")){
            node = new TreeNode(Integer.valueOf(strr[index]));
            node.left = Deserialize(str);
            node.right = Deserialize(str);
        }

        return node;
  }
}

原文地址：https://www.cnblogs.com/muche-moqi/p/12393042.html