Verify Preorder Serialization Of A Binary Tree

                                            
class Solution {
public:
    bool isValidSerialization(string preorder) {
        stack<string> nodes;
        stringstream ss(preorder);
        string node;
        
        while (getline(ss, node, ',')) {
            while (!nodes.empty() && node == "#" && nodes.top() == "#") {
                nodes.pop();
                
                if (nodes.empty()) {
                    return false;
                }
                
                nodes.pop();
            }
            
            nodes.push(node);
        }
        
        return nodes.size() == 1 && nodes.top() == "#";
    }
};

                                            
class Solution {
    public boolean isValidSerialization(String preorder) {
        int slots = 1;
        String[] nodes = preorder.split(",");
        
        for (String node : nodes) {
            slots--;
            
            if (slots < 0) {
                return false;
            }
            
            if (!node.equals("#")) {
                slots += 2;
            }
        }
        
        return slots == 0;
    }
}

                                            
class Solution(object):
    def isValidSerialization(self, preorder):
        """
        :type preorder: str
        :rtype: bool
        """
        stack = []
        for node in preorder.split(","):
            stack.append(node)
            while len(stack) >= 3 and stack[-1] == stack[-2] == "#" and stack[-3] != "#":
                stack.pop()
                stack.pop()
                stack.pop()
                stack.append("#")
         
        return len(stack) == 1 and stack[0] == "#"

                                            
bool isValidSerialization(char* preorder) {
    int slots = 1;
    char* token = strtok(preorder, ",");
    
    while (token != NULL) {
        if (slots == 0) {
            return false;
        }
        
        if (strcmp(token, "#") == 0) {
            slots--;
        } else {
            slots++;
        }
        
        token = strtok(NULL, ",");
    }
    
    return slots == 0;
}

                                            
/**
 * @param {string} preorder
 * @return {boolean}
 */
const isValidSerialization = (preorder) => {
    const nodes = preorder.split(',');
    let slots = 1;
    
    for (let node of nodes) {
        if (slots <= 0) {
            return false;
        }
        
        if (node === '#') {
            slots--;
        } else {
            slots++;
        }
    }
    
    return slots === 0;
};

                                            
function isValidSerialization(preorder: string): boolean {
    const nodes = preorder.split(",");
    let slots = 1;
    
    for (let node of nodes) {
        slots--;
        
        if (slots < 0) {
            return false;
        }
        
        if (node !== "#") {
            slots += 2;
        }
    }
    
    return slots === 0;
};

                                            
class Solution {
    /**
     * @param String $preorder
     * @return Boolean
     */
    function isValidSerialization($preorder) {
        $slots = 1;
        $nodes = explode(",", $preorder);
        foreach ($nodes as $node) {
            $slots--;
            if ($slots < 0) {
                return false;
            }
            if ($node != "#") {
                $slots += 2;
            }
        }
        return $slots == 0;
    }
}

                                            
class Solution {
    func isValidSerialization(_ preorder: String) -> Bool {
        var slots = 1
        let nodes = preorder.split(separator: ",")
        
        for node in nodes {
            slots -= 1
            if slots < 0 {
                return false
            }
            
            if node != "#" {
                slots += 2
            }
        }
        
        return slots == 0
    }
}

                                            
class Solution {
    fun isValidSerialization(preorder: String): Boolean {
        var slots = 1
        for (node in preorder.split(",")) {
            if (slots == 0) return false
            if (node == "#") {
                slots--
            } else {
                slots++
            }
        }
        return slots == 0
    }
}

                                            
class Solution {
  bool isValidSerialization(String preorder) {
    List<String> nodes = preorder.split(',');
    
    int diff = 1;
    
    for (int i = 0; i < nodes.length; i++) {
      if (--diff < 0) return false;
      
      if (nodes[i] != '#') {
        diff += 2;
      }
    }
    
    return diff == 0;
  }
}

                                            
# @param {String} preorder
# @return {Boolean}
def is_valid_serialization(preorder)
    nodes = preorder.split(',')
    slots = 1
    
    nodes.each do |node|
        slots -= 1
        return false if slots < 0
        
        if node != '#'
            slots += 2
        end
    end
    
    slots == 0
end

                                            
impl Solution {
    pub fn is_valid_serialization(preorder: String) -> bool {
        let mut slots = 1;
        for node in preorder.split(',') {
            if slots == 0 {
                return false;
            }
            if node == "#" {
                slots -= 1;
            } else {
                slots += 1;
            }
        }
        slots == 0
    }
}

To solve this coding challenge, we need to determine if a given preorder serialization of a binary tree is valid. Preorder serialization involves recording the value of nodes as we traverse the tree in preorder (root, left, right), using a sentinel value (typically

'#'

) for null nodes.

Explanation

In order to check the validity of the provided preorder serialization string, we need a way to simulate the process of constructing the tree without actually reconstructing it. We can utilize a stack to keep track of the nodes we have encountered. Here are the detailed steps and explanation of the approach:

Understanding Preorder Serialization :

When visiting a node in a binary tree using preorder traversal, we first visit the root, then recursively the left subtree, and finally the right subtree.
Null nodes in the binary tree are represented using
'#'

in the serialization.

Using a Stack for Validation :

We utilize a stack to keep track of nodes as we process the elements of the serialized string.
The idea is to recognize a complete subtree when we encounter two
'#'

values in the stack followed by a non-
'#'

value. This represents two null children of a node, thereby indicating a leaf node and its completed traversal.

Processing the Serialization String :

Split the input string by commas to process each node value or
'#'

.
Iterate through the split values, and for each value:

Append it to the stack.
Continuously check if the top of the stack contains a pattern indicating a complete subtree (
node, '#', '#'

).
If such a pattern is found, pop the elements and replace them with a single
'#'

to represent the completed subtree.

Final Check :

At the end of the process, a valid serialization should reduce to exactly one
'#'

in the stack.

Here's the pseudocode with comments for clarity:

                                            
Function isValidSerialization(preorder):
# Split the preorder string by commas to process each node and null marker
nodes = split(preorder, ",")
# Initialize an empty stack to assist with the validation process
stack = []

# Iterate over each node in the split preorder list
for node in nodes:
# Append the current node or null marker to the stack
append(stack, node)

# Continuously check for the pattern that indicates a complete subtree
While length(stack) >= 3 AND top(stack) == '#' AND secondToTop(stack) == '#' AND thirdToTop(stack) != '#':
# Remove the complete subtree pattern from the stack
pop(stack)  # Removes the top '#'
pop(stack)  # Removes the second '#'
pop(stack)  # Removes the non-'#' node
# Append a single '#' to represent the completed subtree
append(stack, '#')

# After processing all nodes, a valid serialization will have one '#' in the stack
Return length(stack) == 1 AND top(stack) == '#'

Step-by-Step Explanation:

Initialization :

We start by splitting the preorder string into a list of nodes for easier processing.
Initialize an empty stack to keep track of nodes as we parse through the list.

Processing Each Node :

Iterate through each element in the nodes list:

Append the current node or
'#'

to the stack.
Check if the last three elements in the stack form the pattern of a complete subtree (
node, '#', '#'

).
If the pattern is detected, remove these three elements from the stack and replace them with a single
'#'

to signify the completed subtree.

Final Verification :

After all nodes have been processed, check if the stack has been reduced to a single
'#'

.
If it is, the serialization is valid; otherwise, it is not.

Using this approach ensures we can validate the serialization string efficiently without reconstructing the binary tree. The stack-based approach helps in identifying the completeness of subtrees dynamically, adhering to the properties of tree traversal and serialization.