Code & Func

第104天。

今天的题目是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 #.

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     _9_
2
    /   \
3
   3     2
4
  / \   / \
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 4   1  #  6
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/ \ / \   / \
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# # # #   # #

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

虽然题目很长，但是理解起来并不难，就是给你一串字符串表示一棵二叉树，用,分隔节点的值，用#表示空指针，然后问你这个字符串能不能还原出来一棵二叉树（在不建树的情况下）,其实和建树很像，都是递归的去做：

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bool isValidSerialization(string preorder) {
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    int beg = 0;
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    return isValidSerialization(preorder,beg) && !next(preorder, beg);
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}
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bool isValidSerialization(string preorder, int &beg) {
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    if (beg >= preorder.size()) return false;
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    if (preorder[beg] == '#') return true;
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    return next(preorder, beg) && isValidSerialization(preorder, beg) &&
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        next(preorder, beg) && isValidSerialization(preorder, beg);
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}
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bool next(string &preorder, int &beg) {
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    while(beg < preorder.size() && preorder[beg] != ',') beg++;
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    beg++;
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    return beg < preorder.size();
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}

然后是dicuss中的迭代版本：

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bool isValidSerialization(string preorder) {
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    if (preorder.empty()) return false;
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    preorder+=',';
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    int sz=preorder.size(),idx=0;
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    int capacity=1;
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    for (idx=0;idx<sz;idx++){
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        if (preorder[idx]!=',') continue;
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        capacity--;
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        if (capacity<0) return false;
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        if (preorder[idx-1]!='#') capacity+=2;
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    }
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    return capacity==0;
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}

Update at 2020-04-02

补充一个自己的写法：

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bool isValidSerialization(string preorder) {
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    int i, c, size;
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    for(i = 0, c = 1, size = preorder.size(); i < size && c; i++){
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        if (preorder[i] == '#') c--;
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        else c++; // c--; c+=2;
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        // move next node
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        while(i < size && preorder[i]!=',') i++;
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    }
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    return c == 0 && i >= size;
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}