Question

empty tree ::= ()
tree ::= empty tree | (w tree tree)
ex:
()
empty tree

(99(5()())(35(-5()())()))

     99
    /  \
   5   35
      /
     -5

class Node
{
public:
    int weight; // weight can be negative!
    Node *left, *right;
    Node():weight(0),left(NULL),right(NULL){}
    Node(int d):weight(d),left(NULL),right(NULL){}
};

按给定条件

构造二叉树

我遇到构造它的问题，我的程序会粉碎，我不知道它为什么会发生，以下是我的代码，我打印出一些信息进行调试，取（99（5（）（））（35 （-5（）（））（）））作为一个测试用例，它将打印出99（5（并且粉碎，我想也许是我处理的问题）我返回的节点是NULL，但我可以找不到它的问题。顺便说一句，这棵树应该处理每棵树中的HUNDREDS节点，并且每个测试用例包含多达TEN-THOUSAND树，我会用这个程序耗尽时间或者应该用什么我需要做什么？感谢你的时间

Node* MyBinaryTreeOps::constructTree(Node *root, std::string treeStr)const
{
    int idex = 1;//always look at the treeStr[1]
    Node *cur=NULL;//use to pass in recursive call
    if(treeStr[idex]!='('&&treeStr[idex]!=')'){//meet number create new node
       stringstream ss;
       while(treeStr[idex]!='('){
             ss<<treeStr[idex];
             if(treeStr.size()>1){//if size > 1 then remove the treeStr[1],to let treeStr[1] become next char in treeStr
                treeStr.erase(1,1);
             }
        }
        int num=0;
        ss>>num;
        std::cout<<num<<std::endl;//print out just for debug
        std::cout<<treeStr[idex]<<std::endl;//print out just for debug
        root = new Node(num);
     }

    if(treeStr[idex]==')'){//meet ')' return subtree constructed
      if(treeStr.size()>1){
         treeStr.erase(1,1);
      }
       return root;
    }
    if(treeStr[idex]=='('){//meet first '(' then construct left subtree
       if(treeStr.size()>1){
          treeStr.erase(1,1);
       }

       root->left = constructTree(cur,treeStr);

    }

    if(treeStr[idex]=='('){ //meet second '(' then construct right subtree
       if(treeStr.size()>1){
          treeStr.erase(1,1);
       }
       root->right = constructTree(cur,treeStr);

    }
    if(treeStr[idex]==')'){ //meet ')' return subtree constructed
       if(treeStr.size()>1){
          treeStr.erase(1,1);
       }
       return root;
    }
}

Answer 1

我自己尝试过这个问题，这是我写的功能。

算法的步骤：

找到表示当前节点权重的序列的一部分。将其转换为int并分配给节点。
切割线以消除重量，开始和结束支撑。
迭代序列以找到划分子节点的两个括号之间的点。
将子串分为两个序列（我们可以将起始树切片并将其重新用作其中一个子节点的序列）。
如果子节点具有权重（其序列的长度大于2），则创建新节点并递归算法。

此外，这是我的程序，其中包含一些测试示例和一些扩展的Node类：

Node* constructTree(Node* root, std::string& treeString) {
    // Find the weight of this node.
    auto weightLeft = treeString.find_first_of("(") + 1;
    auto weightRight = treeString.find_first_of("()", weightLeft);
    auto weightString = treeString.substr(weightLeft, weightRight - weightLeft);

    // Optional, we check if there is any weight, if there is not we leave zero
    // weight from constructor.
    // Works for something like that: ((1)(2)) -> (0(1)(2))
    if (weightString.length() > 0) {
        root->weight = std::stoi(weightString);
    }

    // Slice string to contain only children sequences.
    treeString.erase(0, weightRight);
    treeString.erase(treeString.length() - 1, 1);

    // Looking for index in string where a left child ends and a right child starts.
    // This point(index) is located where count of left braces and for braces
    // is the same and the counts are not zero.
    int splitPoint = -1;
    int leftBraces = 0, rightBraces = 0;
    for (int index = 0; index < treeString.length(); index++) {
        char c = treeString[index];
        if (c == '(') {
            ++leftBraces;
        }
        if (c == ')') {
            ++rightBraces;
        }

        if (leftBraces == rightBraces) {
            splitPoint = index + 1;
            break;
        }
    }

    // If split point has been found then it means that this node has children.
    if (splitPoint != -1) {
        auto leftChildString = treeString.substr(0, splitPoint);
        auto rightChildString = treeString.erase(0, splitPoint);

        // Check for length so construct will stop if there is no child.
        if (leftChildString.length() > 2) {
            root->left = new Node();
            constructTree(root->left, leftChildString);
        }

        if (rightChildString.length() > 2) {
            root->right = new Node();
            constructTree(root->right, rightChildString);
        }
    }

    return root;
}

从c ++中的s-expression构造二叉树

1 个答案: