这里想到的比较容易的方式是使用层序遍历的结果来序列化,再利用层序遍历的方式来反序列树。
但是需要注意的是,由于我们要序列化的树是唯一的,所以我们需要标记空节点。
并且,我们将序列化的结果按照顺序加入到数组中,查看根节点和子树的关系。
对于某个节点node,它在序列中的位置确定好后,就可以确定好子树的索引。
/**
* 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) {
StringBuilder res = new StringBuilder();
if(root != null) {
Queue q = new ArrayDeque<>();
q.offer(root);
res.append(root.val);
res.append(",");
while(!q.isEmpty()) {
int n = q.size();
for(int i = 0; i < n; i++) {
TreeNode node = q.poll();
if(node.left != null) {
res.append(node.left.val);
res.append(",");
q.offer(node.left);
} else {
res.append("#,");
}
if(node.right != null) {
res.append(node.right.val);
res.append(",");
q.offer(node.right);
} else {
res.append("#,");
}
}
}
}
// System.out.println(Arrays.toString(res.toString().split(",")));
return res.toString();
}
// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
if(data == null || data.length() == 0) {
return null;
}
String[] vals = data.split(",");
if(vals == null || vals.length == 0) {
return null;
} else {
int idx = 0;
Queue q = new ArrayDeque<>();
TreeNode root = new TreeNode(Integer.parseInt(vals[idx]));
q.offer(root);
idx += 1;
while(!q.isEmpty() && idx < vals.length) {
TreeNode node = q.poll();
if(!vals[idx].equals("#")) {
node.left = new TreeNode(Integer.parseInt(vals[idx]));
q.offer(node.left);
} else {
node.left = null;
}
if(idx + 1 < vals.length && !vals[idx +1].equals("#")) {
node.right = new TreeNode(Integer.parseInt(vals[idx + 1]));
q.offer(node.right);
} else {
node.right = null;
}
idx += 2;
}
return root;
}
}
}
// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.deserialize(codec.serialize(root));
Original: https://www.cnblogs.com/nullpointer-c/p/15884599.html
Author: NullPointer_C
Title: 剑指 Offer 37. 序列化二叉树
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