_331.java

package com.fishercoder.solutions;

import java.util.ArrayDeque;
import java.util.Deque;

/**
 * 331. Verify Preorder Serialization of a Binary Tree
 *
 * One way to serialize a binary tree is to use pre-order 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 #.

       _9_
      /    \
     3     2
    / \   / \
   4  1  #  6
  / \ / \  / \
  # # # #  # #

 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, verify whether it is a correct preorder traversal serialization of a binary tree. Find an algorithm without reconstructing the tree.

 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".

 Example 1:
 "9,3,4,#,#,1,#,#,2,#,6,#,#"
 Return true

 Example 2:
 "1,#"
 Return false

 Example 3:
 "9,#,#,1"
 Return false


 */
public class _331 {

    /**credit: https://discuss.leetcode.com/topic/35976/7-lines-easy-java-solution
     * Some used stack. Some used the depth of a stack. Here I use a different perspective. In a binary tree, if we consider null as leaves, then
     all non-null node provides 2 outdegree and 1 indegree (2 children and 1 parent), except root
     all null node provides 0 outdegree and 1 indegree (0 child and 1 parent).
     Suppose we try to build this tree.
     During building, we record the difference between out degree and in degree diff = outdegree - indegree.
     When the next node comes, we then decrease diff by 1, because the node provides an in degree.
     If the node is not null, we increase diff by 2, because it provides two out degrees.
     If a serialization is correct, diff should never be negative and diff will be zero when finished.
     */
    public boolean isValidSerialization_clever_solution(String preorder) {
        String[] pre = preorder.split(",");
        int diff = 1;
        for (String each : pre) {
            if (diff < 0) {
                return false;
            }
            diff--;
            if (!each.equals("#")) {
                diff += 2;
            }
        }
        return diff == 0;
    }

    public boolean isValidSerialization(String preorder) {
        /**Idea: we keep inserting the string into the stack, if it's a number, we just push it onto the stack;
         * if it's a "#", we see if the top of the stack is a "#" or not,
         * 1. if it's a "#", we pop it and keep popping numbers from the stack,
         * 2. if it's not a "#", we push the "#" onto the stack*/
        if (preorder == null || preorder.length() == 0) {
            return false;
        }
        String[] pre = preorder.split(",");
        Deque<String> stack = new ArrayDeque<>();
        for (int i = 0; i < pre.length; i++) {
            while (pre[i].equals("#") && !stack.isEmpty() && stack.peekLast().equals("#")) {
                stack.pollLast();
                if (stack.isEmpty()) {
                    return false;
                }
                stack.pollLast();
            }
            stack.addLast(pre[i]);
        }
        return stack.size() == 1 && stack.peekLast().equals("#");
    }

}