331. Verify Preorder Serialization of a Binary Tree -- 判断是否为合法的先序序列

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

1. 栈

class Solution {
public:
bool isValidSerialization(string preorder) {
stack<char> stk;
bool isNum = false;
preorder.push_back(','); // dummy tail

for(auto c: preorder){
if(c == '#'){
// absorb: search for pattern `#, number` backward
while(!stk.empty() && stk.top() == '#'){
stk.pop(); // pop `#`
if(stk.empty() || stk.top() == '#') return false; // pattern `#,#,#`
stk.pop(); // pop `number`
}
stk.push('#'); // replace `number` with `#` since it has been fully explored/validated
}else if(c == ','){
if(isNum) stk.push('n'); // indicate this is a number instead of using the real number
isNum = false;
}else{
isNum = true;
}
}

return stk.size() == 1 && stk.top() == '#';
}
};

2. 不用栈

class Solution {
public:
bool isValidSerialization(string preorder) {
string& s = preorder;
while (s.size() >= 5) {
bool find_pattern = false;
for (int i = s.size()-1; i>= 4; i--) {
if (s[i] == '#' && s[i-2] == '#' && s[i-4] != '#') {
find_pattern = true;
int j = i-4-1;
/* find the start place of pattern */
while (j > 0 && s[j] != ',') j--;
s.replace(j+1, i-j, "#"); /* replace s[j+1, i]  to "#" */
break;  /* start a trun search from the end */
}
}
if (!find_pattern) break;
}

/* boundary: empty tree */
return (s.size() == 1 && s[0] == '#');
}
};

从右向左，若出现(数字,#,#)模式，则替换成一个#。最后若只剩一个#则合法，否则不合法。