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;
}