The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants.
Given any two nodes in a binary tree, you are supposed to find their LCA.
Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers: M (≤ 1,000), the number of pairs of nodes to be tested; and N (≤ 10,000), the number of keys in the binary tree, respectively. In each of the following two lines, N distinct integers are given as the inorder and preorder traversal sequences of the binary tree, respectively. It is guaranteed that the binary tree can be uniquely determined by the input sequences. Then M lines follow, each contains a pair of integer keys U and V. All the keys are in the range of int.
Output Specification:
For each given pair of U and V, print in a line LCA of U and V is A. if the LCA is found and A is the key. But if A is one of U and V, print X is an ancestor of Y. where X is A and Y is the other node. If U or V is not found in the binary tree, print in a line ERROR: U is not found. or ERROR: V is not found. or ERROR: U and V are not found..
Sample Input:
6 8
7 2 3 4 6 5 1 8
5 3 7 2 6 4 8 1
2 6
8 1
7 9
12 -3
0 8
99 99
Sample Output:
LCA of 2 and 6 is 3.
8 is an ancestor of 1.
ERROR: 9 is not found.
ERROR: 12 and -3 are not found.
ERROR: 0 is not found.
ERROR: 99 and 99 are not found.
有一道考研题也是求lca，突然想到一个不用建树的方法，不过要额外小心细节，避免超时。依然用map记录元素在中序遍历数组中位置，下标不为0，如下：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
using namespace std;
int pre[],in[],m,n;
map<int,int> mp;
void check(int a,int b) {
int aa = mp[a],bb = mp[b];
if(!bb && !aa)printf("ERROR: %d and %d are not found.\n",a,b);
else if(!aa)printf("ERROR: %d is not found.\n",a);
else if(!bb)printf("ERROR: %d is not found.\n",b);
else {
if(a == b) {
printf("%d is an ancestor of %d.\n",a,b);
return;
}
int p1 = ,p2 = n,i1 = ,i2 = n;
while(p1 <= p2) {
int temp = pre[p1],mtemp = mp[temp];
if(temp == a) {
printf("%d is an ancestor of %d.\n",a,b);
return;
}
else if(temp == b) {
printf("%d is an ancestor of %d.\n",b,a);
return;
}
if(mtemp > aa && mtemp > bb) i2 = mtemp - ,p1 ++,p2 = p1 + i2 - i1;
else if(mtemp < aa && mtemp < bb) i1 = mtemp + ,p1 = p2 - i2 + i1;
else {
printf("LCA of %d and %d is %d.\n",a,b,temp);
return;
}
}
}
}
int main() {
int a,b;
scanf("%d%d",&m,&n);
for(int i = ;i <= n;i ++) {
scanf("%d",&in[i]);
mp[in[i]] = i;
}
for(int i = ;i <= n;i ++) {
scanf("%d",&pre[i]);
}
for(int i = ;i < m;i ++) {
scanf("%d%d",&a,&b);
check(a,b);
}
return ;
}
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
using namespace std;
struct tree
{
int Data,Height;
tree *Last,*Left,*Right;
}*head;
int q[],z[],m,n;
map<int,tree *> mp;///根据序号映射到结点
tree *createNode(int d,int h)///创建新结点并返回
{
tree *p = new tree();
p -> Data = d;
mp[d] = p;
p -> Height = h;
p -> Last = p -> Left = p -> Right = NULL;
return p;
}
tree *createTree(int ql,int qr,int zl,int zr,int h)///前中序 还原树
{
tree *p = createNode(q[ql],h);
for(int i = zl;i <= zr;i ++)
{
if(z[i] == q[ql])
{
if(i > zl)p -> Left = createTree(ql + ,ql + i - zl,zl,i - ,h + ),p -> Left -> Last = p;
if(i < zr)p -> Right = createTree(ql + i - zl + ,qr,i + ,zr,h + ),p -> Right -> Last = p;
break;
}
}
return p;
}
void check(int a,int b)///判断两点
{
if(mp[a] == NULL && mp[b] == NULL)printf("ERROR: %d and %d are not found.\n",a,b);
else if(mp[a] == NULL)printf("ERROR: %d is not found.\n",a);
else if(mp[b] == NULL)printf("ERROR: %d is not found.\n",b);
else {
tree *t1 = mp[a],*t2 = mp[b];
while(t1 -> Height != t2 -> Height) {
if(t1 -> Height > t2 -> Height)t1 = t1 -> Last;
else t2 = t2 -> Last;
}///调整至高度相同
if(t1 == t2) {///一个点是另一个点的祖先
printf("%d is an ancestor of %d.\n",t1 -> Data,a == t1 -> Data ? b : a);
return;
}while(t1 != t2) {
t1 = t1 -> Last;
t2 = t2 -> Last;
}
printf("LCA of %d and %d is %d.\n",a,b,t1 -> Data);
}
}
int main() {
int a,b;
scanf("%d%d",&m,&n);
for(int i = ;i < n;i ++) {
scanf("%d",&z[i]);
}
for(int i = ;i < n;i ++) {
scanf("%d",&q[i]);
}
head = createTree(,n - ,,n - ,);
for(int i = ;i < m;i ++) {
scanf("%d%d",&a,&b);
check(a,b);
}
}