解决思路:
- 按照一棵树的出度跟入度相等的规则来做。首先这棵树加上了“#”,那么将“#”也作为二叉树的一部分来算。根节点只有出度,将diff=1.
class Solution {
public boolean isValidSerialization(String preorder) {
String[] str = preorder.split(",");
int diff = 1;
for(int i = 0; i < str.length; i++){
if(--diff < 0){
return false;
}
if(!str[i].equals("#")){
diff += 2;
}
}
return diff == 0;
}
}