Algorithm
Problem Name: 331. Verify Preorder Serialization of a Binary Tree
One way to serialize a binary tree is to use preorder traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record using a sentinel value such as '#'.
For example, the above binary tree can be serialized to the string "9,3,4,#,#,1,#,#,2,#,6,#,#", where '#' represents a null node.
Given a string of comma-separated values preorder, return true if it is a correct preorder traversal serialization of a binary tree.
It is guaranteed that each comma-separated value in the string must be either an integer or a character '#' representing null pointer.
You may assume that the input format is always valid.
- For example, it could never contain two consecutive commas, such as
"1,,3".
Note: You are not allowed to reconstruct the tree.
Example 1:
Input: preorder = "9,3,4,#,#,1,#,#,2,#,6,#,#" Output: true
Example 2:
Input: preorder = "1,#" Output: false
Example 3:
Input: preorder = "9,#,#,1" Output: false
Constraints:
1 <= preorder.length <= 104preorderconsist of integers in the range[0, 100]and'#'separated by commas','.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
bool isValidSerialization(string preorder) {
if(preorder == "#") return true;
stringstream ss(preorder);
stack < int>stk;
string s = "";
while(getline(ss, s, ',')){
if(s == "#" && stk.empty()) return false;
if(!stk.empty()) stk.top()++;
if(!stk.empty() && stk.top() == 2) stk.pop();
if(s != "#") stk.push(0);
if(stk.empty() && !ss.eof()) return false;
}
return stk.empty();
}
};
Input
Output
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public boolean isValidSerialization(String preorder) {
String[] nodes = preorder.split(",");
int diff = 1;
for (String node : nodes) {
if (--diff < 0) {
return false;
}
if (!node.equals("#")) {
diff += 2;
}
}
return diff == 0;
}
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const isValidSerialization = function(preorder) {
const nodes = preorder.split(',')
let diff = 1
for(let node of nodes) {
if(--diff < 0) return false
if(node !== '#'> diff += 2
}
return diff === 0
};
Input
Output
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def isValidSerialization(self, preorder: str) -> bool:
stack = []
for c in preorder.split(','):
stack.append(c)
while stack[-2:] == ['#', '#']:
stack.pop()
stack.pop()
if not stack: return False
stack.pop()
stack.append('#')
return stack == ['#']
Input
Output