Verify Preorder Serialization of a Binary Tree

Description

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

Hint

Train of Thought

Stack: 碰到连续两个#时说明该节点是叶子节点, 用一个#来代替它, 继续往上走, 4 和 1都被#代替后, 3就变成了叶子节点, 这样层层向上消去节点, 直到最后的root节点也可以用#表示.

Graph: 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.

Code

//Stack
public class Solution {
    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("#");
    }
}

//Graph
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;

}

Complexity