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题目描述

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.

A binary search tree (BST) is recursively defined as a binary tree which has the following properties:

The left subtree of a node contains only nodes with keys less than the node’s key.

The right subtree of a node contains only nodes with keys greater than or equal to the node’s key. Both the left and right subtrees must also be binary search trees. Given any two nodes in a BST, 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 BST, respectively. In the second line, N distinct integers are given as the preorder traversal sequence of the BST. 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 BST, 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 6 3 1 2 5 4 8 7 2 5 8 7 1 9 12 -3 0 8 99 99 Sample Output: LCA of 2 and 5 is 3. 8 is an ancestor of 7. ERROR: 9 is not found. ERROR: 12 and -3 are not found. ERROR: 0 is not found. ERROR: 99 and 99 are not found.

题意

给出一颗二叉搜索树的前序遍历序列，求任意两个节点的LCA。

思路

直接通过前序序列去建立二叉搜索树会超时，因为极端的情况下的时间复杂度会达到N^2。

所以我们可以通过前序和中序序列去建造树，二叉搜索树的中序序列其实就是前序序列升序排序的序列。 建完树之后，就需要找LCA了。 找LCA我是用的暴力的方法，即先将两个节点调到同一高度，之后再共同向上找LCA。 注意特殊情况，如果x等于y的话，则输出x is an ancestor of x 这样的形式。

代码如下

#include 
   
    using namespace std;const int maxm=1005;const int maxn=10005;int n,m;struct tree{
       int left,right;    int fa;    int dep;    int val;};int pre[maxn];int in[maxn];tree t[maxn];int num=0;int root=-1;map
    
     ma;int build(int pre_left,int pre_right,int in_left,int in_right,int fa,int dep){
       if(pre_left>pre_right) return -1;    //printf("pre=%d in=%d\n",pre_right-pre_left,in_right-in_left);    int val=pre[pre_left];    ma[val]=num;    t[num].val=val;    t[num].fa=fa;    t[num].dep=dep;    int loc=-1;    int tnum=num;    for (int i=in_left;i<=in_right;i++)    {
           if(in[i]==val)        {
               loc=i;            break;        }    }    num++;    t[tnum].left=build(pre_left+1,pre_left+loc-in_left,in_left,loc-1,tnum,dep+1);    t[tnum].right=build(pre_left+loc-in_left+1,pre_right,loc+1,in_right,tnum,dep+1);    return tnum;}//调到同一高度后找lcavoid lca(int inx,int iny,int x,int y){
        while(inx!=iny)    {
           inx=t[inx].fa;        iny=t[iny].fa;    }    printf("LCA of %d and %d is %d.\n",x,y,t[inx].val);}//使两个节点处在同一高度int reach_same(int inx,int iny){
       while(t[inx].dep>t[iny].dep)    {
           inx=t[inx].fa;    }    return inx;}int main(){
       memset (t,-1,sizeof(t));    scanf("%d%d",&m,&n);    for (int i=0;i
     
      dy)                {
                       inx=reach_same(inx,iny);                    if(inx==iny)                    {
                           printf("%d is an ancestor of %d.\n",y,x);                    }                    else                    {
                           lca(inx,iny,x,y);                    }                }                else                {
                       iny=reach_same(iny,inx);                    if(inx==iny)                    {
                           printf("%d is an ancestor of %d.\n",x,y);                    }                    else                    {
                           lca(inx,iny,x,y);                    }                }            }        }    }    return 0;}
     
    
   

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