题解 | #序列化二叉树#

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

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

}

} / import java.util.;

public class Solution {

//一个简单的BFS广度优先遍历
String Serialize(TreeNode root) {

    if(root == null){
        return "";
    }
    int depth = getDepth(root);
    int size = (int)Math.pow(2,depth) * 2 - 3;
    String str = "";
    Queue<TreeNode> que = new LinkedList<TreeNode>();
    que.offer(root);
    while(!que.isEmpty()){
        TreeNode tmpNode = que.poll();
        if(tmpNode != null){
            str  += tmpNode.val;
        }else{
            str += '#';
        }
        str += " ";
        if(tmpNode != null){
            que.offer(tmpNode.left);
            que.offer(tmpNode.right);
        }

    }
    return str;
}

public static int getDepth(TreeNode root){
    if(root == null){
        return 0;
    }
    return Math.max(getDepth(root.left),getDepth(root.right)) + 1;
}


TreeNode Deserialize(String str) {

    if(str.equals("")){return null;}
    //分割成数组形式
    String[] s1 = str.split(" ");
    int len = s1.length;//数组长度
    
    //逆过程
    Queue<TreeNode> que1 = new LinkedList<TreeNode>();
    //头节点
    TreeNode root = new TreeNode(Integer.parseInt(s1[0]));
    que1.offer(root);
    
    for(int i = 1; i < s1.length - 1; i += 2){
    //注意：两个数一组，因为要给倒出节点的左右节点分别赋值
        TreeNode tmpNode = que1.poll();
        //s1[i]非“#”，则s1[i]给左节点赋值，否则左节点为null
        if(!s1[i].equals("#")){
            int l = Integer.parseInt(s1[i]);
            tmpNode.left = new TreeNode(l);
            que1.offer(tmpNode.left);
        }
        //同上
        if(!s1[i + 1].equals("#")){
            int r = Integer.parseInt(s1[i + 1]);
            tmpNode.right = new TreeNode(r);
            que1.offer(tmpNode.right);
        }
    } 
    return root;
}

}