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
分析:https://www.hrwhisper.me/leetcode-verify-preorder-serialization-of-a-binary-tree/
这个方法简单的说就是不断的砍掉叶子节点。最后看看能不能全部砍掉。
以例子一为例,:”9,3,4,#,#,1,#,#,2,#,6,#,#” 遇到x # #的时候,就把它变为 #
我模拟一遍过程:
- 9,3,4,#,# => 9,3,# 继续读
- 9,3,#,1,#,# => 9,3,#,# => 9,# 继续读
- 9,#2,#,6,#,# => 9,#,2,#,# => 9,#,# => #
1 public boolean isValidSerialization(String preorder) { 2 Stack<String> stack = new Stack<String>(); 3 String[] arr = preorder.split(","); 4 5 for (int i = 0; i < arr.length; i++) { 6 stack.push(arr[i]); 7 while (stack.size() >= 3 && stack.get(stack.size() - 1).equals("#") 8 && stack.get(stack.size() - 2).equals("#") && !stack.get(stack.size() - 3).equals("#")) { 9 stack.remove(stack.size() - 2); 10 stack.remove(stack.size() - 2); 11 } 12 } 13 14 if (stack.size() == 1 && stack.peek().equals("#")) 15 return true; 16 17 return false; 18 }
另一种解法:https://www.hrwhisper.me/leetcode-verify-preorder-serialization-of-a-binary-tree/
看了 dietpepsi 的代码,确实思路比我上面的更胜一筹:
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 decreasediff
by 1, because the node provides an in degree. If the node is notnull
, we increase diff by2
, because it provides two out degrees. If a serialization is correct, diff should never be negative and diff will be zero when finished.from :https://leetcode.com/discuss/83824/7-lines-easy-java-solution
我这里翻译一下:
对于二叉树,我们把空的地方也作为叶子节点(如题目中的#),那么有
- 所有的非空节点提供2个出度和1个入度(根除外)
- 所有的空节点但提供0个出度和1个入度
我们在遍历的时候,计算diff = outdegree – indegree. 当一个节点出现的时候,diff – 1,因为它提供一个入度;当节点不是#的时候,diff+2(提供两个出度) 如果序列式合法的,那么遍历过程中diff >=0 且最后结果为0.
1 public boolean isValidSerialization(String preorder) { 2 String[] nodes = preorder.split(","); 3 int diff = 1; 4 for (String node: nodes) { 5 if (--diff < 0) return false; 6 if (!node.equals("#")) diff += 2; 7 } 8 return diff == 0; 9 }