剑指offer 61.树 序列化二叉树

题目描述

请实现两个函数,分别用来序列化和反序列化二叉树
 
二叉树的序列化是指:把一棵二叉树按照某种遍历方式的结果以某种格式保存为字符串,从而使得内存中建立起来的二叉树可以持久保存。序列化可以基于先序、中序、后序、层序的二叉树遍历方式来进行修改,序列化的结果是一个字符串,序列化时通过 某种符号表示空节点(#),以 ! 表示一个结点值的结束(value!)。

二叉树的反序列化是指:根据某种遍历顺序得到的序列化字符串结果str,重构二叉树。
 

代码如下

//采用层序遍历,不需要将转化为完全二叉树的简单方法
public class Solution {
    String Serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        if(root != null)
            queue.add(root);
        while(!queue.isEmpty()){
            TreeNode node = queue.poll();
            if(node != null){
                queue.offer(node.left);
                queue.offer(node.right);
                sb.append(node.val + ",");
            }else{
                sb.append("#" + ",");
            }
        }
        if(sb.length() != 0)
            sb.deleteCharAt(sb.length()-1);
        return sb.toString();
  }
    TreeNode Deserialize(String str) {
       TreeNode head = null;
        if(str == null || str.length() == 0)
            return head;
        String[] nodes = str.split(",");
        TreeNode[] treeNodes = new TreeNode[nodes.length];
        for(int i=0; i<nodes.length; i++){
            if(!nodes[i].equals("#"))
                treeNodes[i] = new TreeNode(Integer.valueOf(nodes[i]));
        }
        for(int i=0, j=1; j<treeNodes.length; i++){
            if(treeNodes[i] != null){
                treeNodes[i].left = treeNodes[j++];
                treeNodes[i].right = treeNodes[j++];
            }
        }
        return treeNodes[0];
  }
}
 
//前序遍历
public class Solution {
    String Serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        getSerializeString(root, sb);
        if(sb.length() != 0)
            sb.deleteCharAt(sb.length()-1);
        return sb.toString();
      }
    getSerializeString(TreeNode root, StringBuilder sb){
        if(root == null)
            sb.append("#,");
        else{
            sb.append(root.val + ",");
            getSerializeString(root.left, sb);
            getSerializeString(root.right, sb);
        }
    }
    TreeNode Deserialize(String str) {
       if(str == null || str.length() == 0 || str.length() ==1)
            return null;
        String[] nodes = str.split(",");
        TreeNode[] treeNodes = new TreeNode[nodes.length];
        for(int i=0; i<nodes.length; i++){
            if(!nodes[i].equals("#"))
                treeNodes[i] = new TreeNode(Integer.valueOf(nodes[i]));
        }
        Stack<TreeNode> stack = new Stack<>();
        stack.push(treeNodes[0]);
        int i = 1;
        while(treeNodes[i] != null){
            stack.peek().left = treeNodes[i];
            stack.push(treeNodes[i++]);
        }
        while(!stack.isEmpty()){
            stack.pop().right = treeNodes[++i];
            if(treeNodes[i] != null){
                stack.push(treeNodes[i++]);
                while(treeNodes[i] != null){
                    stack.peek().left = treeNodes[i];
                    stack.push(treeNodes[i++]);
                }
            }
        }
        return treeNodes[0];
      }
}

 

 

 

posted @ 2019-08-28 09:57  Transkai  阅读(139)  评论(0编辑  收藏  举报