厦门市app开发网站建设公司/网站推广的基本方法有

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; 
}