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

Method1：

indegree and outdegree.

public class Solution {    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;            }}

总结：利用出度和入度。关键在于二叉树出度和入度相等。初始化时要注意，diff为1，这是为了满足遍历根节点时其实未减少入度但减去了一个入度。

Method2:

Stack:

when you iterate through the preorder traversal string, for each char:

case 1: you see a number c, means you begin to expand a new tree rooted with c, you push it to stack

case 2.1: you see a #, while top of stack is a number, you know this # is a left null child, put it there as a mark for next coming node k to know it is being the right child.

case 2.2: you see a #, while top of stack is #, you know you meet this # as right null child, you now cancel the sub tree (rooted as t, for example) with these two-# children. But wait, after the cancellation, you continue to check top of stack is whether # or a number:

---- if a number, say u, you know you just cancelled a node t which is left child of u. You need to leave a # mark to the top of stack. So that the next node know it is a right child.

---- if a #, you know you just cancelled a tree whose root, t, is the right child of u. So you continue to cancel sub tree of u, and the process goes on and on.

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("#");    }}