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The idea is to track available slots in the binary tree. Initially, we start with one slot for the root. For each node, we consume a slot. If the node is non-null, it adds two more slots (one for each child). A null node ('#') does not add any slots. The binary tree is valid if, at the end of processing all nodes, all slots are exactly filled (i.e., zero slots remaining).
Time Complexity: O(n), where n is the length of the preorder string, as we are iterating through the nodes once.
Space Complexity: O(1), with no additional data structures used.
1public class Solution {
2 public boolean isValidSerialization(String preorder) {
3 int slots = 1;
4 String[] nodes = preorder.split(",");
5 for (String node : nodes) {
6 slots--;
7 if (slots < 0) return false;
8 if (!node.equals("#")) slots += 2;
9 }
10 return slots == 0;
11 }
12 public static void main(String[] args) {
13 Solution solution = new Solution();
14 String preorder = "9,3,4,#,#,1,#,#,2,#,6,#,#";
15 System.out.println(solution.isValidSerialization(preorder));
16 }
17}The Java solution utilizes the split function to tokenize the string by commas. The algorithm then follows the same slot accounting logic to validate the serialization.
This approach involves simulating a stack-based tree traversal. The idea is to treat each "open bracket" as requiring nodes and each "close bracket" as completing them. We aim to manage the state of openings and closures using integers, representing how many nodes are expected at any point in the traversal.
Time Complexity: O(n), parsing each node once.
Space Complexity: O(1), with only integer variables used for tracking.
The JavaScript solution splits the string and applies state transitions to verify binary tree serialization validity.