Leetcode: Verify Preorder Serialization of a Binary Tree

One way to serialize a binary tree is to use pre-oder 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

跟G面经：Valid Preorder traversal serialized string一样

用stack, 注意插入“#”是while

public class Solution {
    public boolean isValidSerialization(String preorder) {
        if (preorder==null || preorder.length()==0) return false;
        String[] strs = preorder.split(",");
        int depth = 0;
        for (int i=0; i<strs.length; i++) {
            String cur = strs[i];
            while (cur.equals("#") && st.size()>1 && st.peek().equals("#")) {
                st.pop();
                st.pop();
            }
            st.push(cur);
        }
        if (st.size()==1 && st.peek().equals("#")) return true;
        return false;
    }
}

有人提供了O(1) space不用stack用两个pointer的做法，还不理解

public class Solution {
    public boolean isValidSerialization(String preorder) {
        if (preorder == null || preorder.length() == 0) return false;
        String[] strs = preorder.split(",");
        int depth = 0;
        int i = 0; 
        while (i < strs.length - 1) {
            if (strs[i++].equals("#")) {
                if (depth == 0) return false;
                else depth--;
            }
            else depth++;
        }
        if (depth != 0) return false;
        return strs[strs.length - 1].equals("#");
    }
}