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。(代码待勘误)

#include <iostream>

#include <string>

#include <list>

using namespace std;

typedef struct Node

{

    int v;     // In this code, default positive Integer.

    Node *child[3];

    Node(int 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;

    int r;

    cin >> r;

    if(r != 0)         // equal to if(!r) return;

    {

        pRoot = new Node(r);

        pRoot->child[0] = createTree();

        pRoot->child[1] = createTree();

        pRoot->child[2] = 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";

    cout << root->v << endl << s;

    printTree(root->child[0], level+1);

    cout << endl << s;

    printTree(root->child[1], level+1);

    cout << endl << s;

    printTree(root->child[2], level+1);

}
void releaseTree(Tree *root){

    if(root == NULL) return;

    releaseTree(root->child[0]);

    releaseTree(root->child[1]);

    releaseTree(root->child[2]);

    delete[] root;

    root = NULL;

}

/******************************************************************/

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)

{

    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[1]->child[1]->child[2];

        Node *node2 = pRoot->child[1]->child[3]->child[1];

        Node *father;

        father = getFirstCommonFather(pRoot, node1, node2);

        cout << father->v << endl;

        releaseTree(pRoot);

    }

    return 0;

}

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