LeetCode OJ 331. Verify Preorder Serialization of a Binary Tree

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

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

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解答

这题好坑啊，逗号也要读取。用到的方法是如果是先序遍历的结果的话遍历结束后的数组NULL节点数一定比非空节点恰好多1。而这里要注意的是非空节点的键值可能在字符串数组中是多位数，并且如果字符串最后以数结尾的话，要注意此时是没有逗号的，所以不能简单的以判断逗号作为一个多位数结束的标志。

bool isValidSerialization(char* preorder) {
    int num = 0, ch = 0, i, j;
    for(i = 0; preorder[i] != 0; i = i + 2){
        if(‘#‘ != preorder[i]){
            num++;
            for(j = i; preorder[j] != ‘,‘&&preorder[j] != 0; j++);
            if(preorder[j] == 0)
                break;
            i = j - 1;
        }
        else{
            ch++;
        }
        if(num + 1 == ch){
            break;
        }
    }
    if(preorder[i + 1] == 0&&num + 1 == ch){
        return true;
    }
    else{
        return false;
    }
}