序列化和反序列化二叉树（java版）

【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

【解题思路1】

//1. 利用前序遍历序列化二叉树，然后从记录的字符串中反序列化二叉树。

//2. 遇到空节点需要用特殊字符加以标记。如“#”

public class Solution {
public int index = -1;
String Serialize(TreeNode root) {
StringBuilder s = new StringBuilder();
if(root == null){
s.append("#,");
return s.toString();
}
s.append(root.val+",");
s.append(Serialize(root.left));
s.append(Serialize(root.right));
return s.toString();
}
TreeNode Deserialize(String str) {
index++;
String[] DLRseq = str.split(",");
TreeNode leave = null;
if(!DLRseq[index].equals("#")){
leave = new TreeNode(Integer.valueOf(DLRseq[index]));
leave.left = Deserialize(str);
leave.right = Deserialize(str);
}
return leave;
}
}

【解题思路2】

//1. 层序遍历二叉树，序列化二叉树

//2. 空节点需要使用特殊字符记录。将二叉树序列化成为一颗完全二叉树，然后反序列化二叉树。

public class Serialize {
String Serialize(TreeNode root) {
if(root == null){
return null;
}
StringBuffer sb = new StringBuffer();
ArrayList<TreeNode> list = new ArrayList<TreeNode>();
int count = (1 << treeDepth(root)) - 1;//计数，拿到此树的深度后计算对应完全二叉树节点数
list.add(root);
count--;
TreeNode tmpNode = null;

//层次遍历二叉树，开始序列化
while(list.size() > 0 && count >= 0){
tmpNode = list.remove(0);
if(tmpNode != null){
sb.append(tmpNode.val+",");
list.add(tmpNode.left);
list.add(tmpNode.right);
}else{
sb.append("#,");//#作为空节点占位符
list.add(null);
list.add(null);
}
count --;
}
return sb.toString();
}

TreeNode Deserialize(String str) {
if(str == null || str.length() == 0){
return null;
}
return Deserialize(str.split(","), 0);
}

TreeNode Deserialize(String[] strings, int index){
TreeNode newNode = null;
if(index < strings.length){
if(!strings[index].equals("#")){
newNode = new TreeNode(Integer.parseInt(strings[index]));
newNode.left = Deserialize(strings, 2 * index + 1);
newNode.right = Deserialize(strings, 2 * index + 2);
}
}
return newNode;
}

int treeDepth(TreeNode root){
int depth = 0;
if(root == null){
return depth;
}else{
int lDepth = treeDepth(root.left) + 1;
int rDepth = treeDepth(root.right) + 1;
return lDepth > rDepth ? lDepth : rDepth;
}
}
}