Chapter7: question 49 - 50

49. 把字符串转换为整数

很多细节需要注意。（空格，符号，溢出等）

Go: 8. String to Integer (atoi)

50. 树种两个结点的最低公共祖先

A. 若是二叉搜索树，直接与根结点对比。 若都大于根节点，则在友子树；若都小于根节点，则在左子树；若根节点介于两数之间，则根节点即为答案。

B. 普通树，若是孩子节点有指向父节点的指针，则问题变为求两个链表的第一个公共结点。 如：37题。

C. 普通树：思路1，若一个结点的子树同时包含两个结点，而它的任一孩子结点的子树却不能同时包含，则该节点即为答案。需要重复遍历，时间复杂度较高。

思路2：先序优先遍历，分别记录从根节点到两个结点的路径。然后转换为求第一个公共结点问题。

#include <iostream>
#include <string>
#include <list>
using namespace std; 

typedef struct Node
{
    char v;     // In this code, default positive Integer.
    Node *child[3];
    Node(char x) : v(x){ child[0] = NULL; child[1] = NULL;child[2] = NULL; }
} Tree;
typedef list<Node*> PATH;
/********************************************************/
/*****        Basic functions  for tree     ***********/
Tree* createTree() // input a preOrder sequence, 0 denote empty node.
{
    Node *pRoot = NULL;
    char r;
    cin >> r;
    if(r != '0')         // equal to if(!r) return;
    {
        pRoot = new Node(r);
        for(int i = 0; i < 3; ++i)
            pRoot->child[i] = createTree();
    }
    return pRoot;
}
void printTree(Tree *root, int level = 1){
    if(root == NULL) { cout << "NULL"; return; };
    string s;
    for(int i = 0; i < level; ++i) s += "\t";
    printf("%c", root->v);
    for(int i = 0; i < 3; ++i)
    {
        cout << endl << s;
        printTree(root->child[i], level+1);
    }
}
void releaseTree(Tree *root){
    if(root == NULL) return;
    for(int i = 0; i < 3; ++i)
        releaseTree(root->child[i]);
    delete[] root;
    root = NULL;
}
/******************************************************************/
/****              获取第一个公共父节点              ******/
bool getPath(Tree *root, Node *node, PATH& path)
{
    if(root == NULL) return false;
    path.push_back(root);
    if(root == node)
        return true;
    bool found = false;
    for(int i = 0; i < 3; ++i)
    {
        found = getPath(root->child[i], node, path);
        if(found) break;
    }
    if(!found) path.pop_back();
    return found;
}

Node* getLastCommonNode(const PATH &path1, const PATH &path2) //  get the last common node of two lists
{
    Node *father = NULL;
    for(auto it1 = path1.begin(), it2 = path2.begin(); it1 != path1.end() && it2 != path2.end(); ++it1, ++it2)
    {
        if(*it1 == *it2) father = *it1;
        else break;
    }
    return father;
}

Node* getFirstCommonFather(Tree *root, Node *node1, Node *node2)
{
    if(root == NULL || node1 == NULL || node2 == NULL) return NULL;
    PATH path1, path2;
    if(getPath(root, node1, path1) && getPath(root, node2, path2))
        return getLastCommonNode(path1, path2);
    return NULL;
}
/************************************************************************/
int main(){
    int TestTime = 3, k = 1;
    while(k <= TestTime)
    {
        cout << "Test " << k++ << ":" << endl; 

        cout << "Create a tree: " << endl;
        Node *pRoot = createTree();
        printTree(pRoot);
        cout << endl; 

        Node *node1 = pRoot->child[0]->child[0]->child[1];
        Node *node2 = pRoot->child[0]->child[2]->child[0];
        Node *father;

        father = getFirstCommonFather(pRoot, node1, node2);
        cout << "the first common father node: " << father->v << endl;
        releaseTree(pRoot);
    }
    return 0;
}