331. Verify Preorder Serialization of a Binary Tree

Leetcode Stack

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:

Input: "9,3,4,#,#,1,#,#,2,#,6,#,#"
Output: true

Example 2:

Input: "1,#"
Output: false

Example 3:

Input: "9,#,#,1"
Output: false

Solution 1

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.

为什么开始为1呢？其实可以这么理解，在根结点上加一个虚拟的根节点。

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;
}

Solution2

 public boolean isValidSerialization(String preorder) {
    // using a stack, scan left to right
    // case 1: we see a number, just push it to the stack
    // case 2: we see #, check if the top of stack is also #
    // if so, pop #, pop the number in a while loop, until top of stack is not #
    // if not, push it to stack
    // in the end, check if stack size is 1, and stack top is #
    if (preorder == null) {
        return false;
    }
    Stack<String> st = new Stack<>();
    String[] strs = preorder.split(",");
    for (int pos = 0; pos < strs.length; pos++) {
        String curr = strs[pos];
        while (curr.equals("#") && !st.isEmpty() && st.peek().equals(curr)) {
            st.pop();
            if (st.isEmpty()) {
                return false;
            }
            st.pop();
        }
        st.push(curr);
    }
    return st.size() == 1 && st.peek().equals("#");
}