题目描述
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
解答:
public static boolean isValidSerialization(String preorder) {
List<String> stack = new ArrayList<String>();
String[] arr = preorder.split(",");
for (int i = 0; i < arr.length; i++) {
stack.add(arr[i]);
while (stack.size() >= 3 && stack.get(stack.size() - 1).equals("#")
&& stack.get(stack.size() - 2).equals("#")
&& !stack.get(stack.size() - 3).equals("#")) {
stack.remove(stack.size() - 1);
stack.remove(stack.size() - 1);
stack.remove(stack.size() - 1);
stack.add("#");
}
}
if (stack.size() == 1 && stack.get(0).equals("#")) return true;
else return false;
}
解法二: public static boolean isValidSerialization(String preorder) {
String[] nodes = preorder.split(",");
int diff = 1;
for (String node : nodes) {
// 入度减一
if (--diff < 0) return false;
// 出度加二
if (!node.equals("#")) diff += 2;
}
return diff == 0;
}
refer: