AcWing 50. 序列化二叉树
原题链接
困难
作者:
不哭死神
,
2021-09-03 18:20:27
,
所有人可见
,
阅读 262
BFS建树以及读取
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public String serialize(TreeNode root) {
if (root == null) {
return "[]";
}
StringBuilder sb = new StringBuilder("[");
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
TreeNode cur = queue.poll();
if (cur != null) {
sb.append(cur.val).append(",");
queue.add(cur.left);
queue.add(cur.right);
} else {
sb.append("null").append(",");
}
}
sb.deleteCharAt(sb.length() - 1);
sb.append("]");
return sb.toString();
}
public TreeNode deserialize(String data) {
if ("[]".equals(data)) {
return null;
}
//去除中括号,并按照,分割
String[] vals = data.substring(1, data.length() - 1).split(",");
TreeNode root = new TreeNode(Integer.parseInt(vals[0]));
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
int i = 1;
while (!queue.isEmpty()) {
TreeNode cur = queue.poll();
if (!vals[i].equals("null")) {
cur.left = new TreeNode(Integer.parseInt(vals[i]));
queue.add(cur.left);
}
i++;
if (!vals[i].equals("null")) {
cur.right = new TreeNode(Integer.parseInt(vals[i]));
queue.add(cur.right);
}
i++;
}
return root;
}
}