剑指offer之面试题50树中两个结点的最低公共祖先

问题描述：

设计一个算法，找出给定的任意两个结点的公共父结点。

实现代码如下：

方法一、用链表记录遍历的路径，而后比较两个链表。

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <limits.h>
struct TreeNode{
int value;
struct TreeNode *left;
struct TreeNode *right;
};
typedef struct TreeNode Node;

struct List{
Node *value;
struct List *next;
};
typedef struct List List;

bool doFindCommonNode(Node *node,int num,List *listNum){
if(node==NULL)return false;
if(node->value==num){
return true;
}
bool rs =false;
List *temp =(List *)malloc(sizeof(List));
temp->value=node;
temp->next=listNum->next;
listNum->next=temp;
rs = doFindCommonNode(node->left,num,listNum);
if(!rs)
rs = doFindCommonNode(node->right,num,listNum);
if(!rs){
listNum->next=temp->next;
free(temp);
}
return rs;
}

Node *findCommonNode(Node *head,int num1,int num2){
Node *rs=NULL;
List *listNum1 =(List *)malloc(sizeof(List));
List *listNum2 =(List *)malloc(sizeof(List));

listNum1->next=NULL;
listNum2->next=NULL;
bool b1,b2;
if(head!=NULL){
b1 = doFindCommonNode(head,num1,listNum1);
if(num1==num2){
if(listNum1->next==NULL)
return NULL;
else
return listNum1->next->value;
}
b2 = doFindCommonNode(head,num2,listNum2);
if(b1 && b2){
List *longlist=listNum1->next;
List *shortlist=listNum2->next;
List *head1=listNum1->next;
List *head2=listNum2->next;
int count1=0,count2=0;

while(head1!=NULL){
count1++;
head1=head1->next;
}

while(head2!=NULL){
count2++;
head2=head2->next;
}

int dis=count1-count2;
if(count1<count2){
List *temp = longlist;
longlist=shortlist;
shortlist=temp;
dis=count2-count1;
}

int i=0;
for(i=0;i<dis;i++)
longlist=longlist->next;

while(shortlist!=NULL && longlist!=NULL && longlist->value!=shortlist->value){
longlist=longlist->next;
shortlist=shortlist->next;
}
if(longlist!=NULL && shortlist!=NULL){
rs=longlist->value;
}
}
}
return rs;
}

int main(int argc, char *argv[])
{
Node *head=(Node *)malloc(sizeof(Node));
head->value=0;
head->right=NULL;
head->left=(Node *)malloc(sizeof(Node));
head->left->value=1;
head->left->left=(Node *)malloc(sizeof(Node));
head->left->left->value=2;
head->left->left->left=NULL;
head->left->left->right=NULL;
head->left->right=(Node *)malloc(sizeof(Node));
head->left->right->value=3;
head->left->right->left=NULL;
head->left->right->right=NULL;
int num1,num2;
scanf("%d %d",&num1,&num2);
Node *rs = findCommonNode(head,num1,num2);
if(rs!=NULL)
printf("the common node is :%d\n",rs->value);
else
printf("there is no comomn Node!\n");
return 0;
}

方法二、直接判断两个结点是否在当前结点的左右两侧，如在，则该结点就是所求。否则继续遍历。
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <limits.h>
struct TreeNode{
int value;
struct TreeNode *left;
struct TreeNode *right;
};
typedef struct TreeNode Node;

bool isfind = false;

bool doFindCommonNode(Node *node,Node **rs,int num1,int num2){
if(isfind || node==NULL)return false;
if(node->value==num1){
return true;
}
if(node->value==num2){
return true;
}
bool left = doFindCommonNode(node->left,rs,num1,num2);
bool right = doFindCommonNode(node->right,rs,num1,num2);
if(num1==num2 && (left==true || right==true)){
isfind=true;
*rs=node;
return true;
}
if(left==true && right==true){
isfind=true;
*rs=node;
return true;
}else if(left==true || right==true){
return true;
}else{
return false;
}
}

Node *findCommonNode(Node *head,int num1,int num2){
Node **rs =(Node **)malloc(sizeof(Node *));
*rs=NULL;
if(head!=NULL){
if(doFindCommonNode(head,rs,num1,num2) && isfind){
printf("find the Node!\n");
}else{
printf("there is no comomn Node!\n");
}
}
return *rs;
}

int main(int argc, char *argv[])
{
Node *head=(Node *)malloc(sizeof(Node));
head->value=0;
head->right=NULL;
head->left=(Node *)malloc(sizeof(Node));
head->left->value=1;
head->left->left=(Node *)malloc(sizeof(Node));
head->left->left->value=2;
head->left->left->left=NULL;
head->left->left->right=NULL;
head->left->right=(Node *)malloc(sizeof(Node));
head->left->right->value=3;
head->left->right->left=NULL;
head->left->right->right=NULL;
int num1,num2;
scanf("%d %d",&num1,&num2);
Node *rs = findCommonNode(head,num1,num2);
if(rs!=NULL)
printf("the common node is :%d\n",rs->value);
return 0;
}
参考资料：

剑指offer

备注：

转载请注明出处：http://blog.csdn.net/wsyw126/article/details/51386597

作者：WSYW126