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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。
331. Verify Preorder Serialization of a Binary Tree -- 判断是否为合法的先序序列
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原文地址：http://www.cnblogs.com/argenbarbie/p/5418352.html
