Leetcode: Verify Preorder Serialization of a Binary Tree

Posted 2020-06-12

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

 1 public class Solution {
 2     public boolean isValidSerialization(String preorder) {
 3         if (preorder==null || preorder.length()==0) return false;
 4         String[] strs = preorder.split(",");
 5         int depth = 0;
 6         for (int i=0; i<strs.length; i++) {
 7             String cur = strs[i];
 8             while (cur.equals("#") && st.size()>1 && st.peek().equals("#")) {
 9                 st.pop();
10                 st.pop();
11             }
12             st.push(cur);
13         }
14         if (st.size()==1 && st.peek().equals("#")) return true;
15         return false;
16     }
17 }

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

 1 public class Solution {
 2     public boolean isValidSerialization(String preorder) {
 3         if (preorder == null || preorder.length() == 0) return false;
 4         String[] strs = preorder.split(",");
 5         int depth = 0;
 6         int i = 0; 
 7         while (i < strs.length - 1) {
 8             if (strs[i++].equals("#")) {
 9                 if (depth == 0) return false;
10                 else depth--;
11             }
12             else depth++;
13         }
14         if (depth != 0) return false;
15         return strs[strs.length - 1].equals("#");
16     }
17 }