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
class Node{
String val;
boolean left;
public Node(String s){
left = false;
val = s;
}
}
public boolean isValidSerialization(String preorder) {
//use stack to store parent when we hit '#', '#' we pop parent
//when we hit '#', we know parent has gone through one part, if we went left before we pop
if(preorder.equals("#")) return true;
Stack<Node> stack = new Stack<>();
String s = "";
for(int i = 0 ; i < preorder.length(); i++){
char c = preorder.charAt(i);
if(c == ','){
stack.push(new Node(s));
s="";
}else if(c == '#'){
i++;
if(stack.empty()) return false;
while(!stack.empty() && stack.peek().left){
stack.pop();
}
if(!stack.empty())
stack.peek().left = true;
if(stack.empty() && i < preorder.length()-1)
return false;
}else{
s += c;
}
}
return stack.empty()&&s.equals("");
}
存‘#’当作flag表示左边已经走完
public class Solution {
public boolean isValidSerialization(String preorder) {
if (preorder == null) {
return false;
}
Stack<String> st = new Stack<>();
String[] strs = preorder.split(",");
for (int pos = 0; pos < strs.length; pos++) {
String curr = strs[pos];
while (curr.equals("#") && !st.isEmpty() && st.peek().equals(curr)) {
st.pop();
if (st.isEmpty()) {
return false;
}
st.pop();
}
st.push(curr);
}
return st.size() == 1 && st.peek().equals("#");
}
}
NO Stack
https://discuss.leetcode.com/topic/35976/7-lines-easy-java-solution/2
Some used stack. Some used the depth of a stack. Here I use a different perspective. 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 decrease diff by 1, because the node provides an in degree. If the node is notnull, we increase diff by 2, because it provides two out degrees. If a serialization is correct, diff should never be negative and diff will be zero when finished.
public boolean isValidSerialization(String preorder) {
String[] nodes = preorder.split(",");
int diff = 1;
for (String node: nodes) {
if (--diff < 0) return false;
if (!node.equals("#")) diff += 2;
}
return diff == 0;
}
