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 }