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.
题意 给出二叉树的中序和前序，再给出m个询问，问这两个节点a,b的最低级共同父节点,并输出LCA of a and b is x. 如果a就是b的父节点，输出a is an ancestor of b. 如果a不存在，输出ERROR: a is not found. 如果都不存在，输出ERROR: a and b are not found.
#include<bits/stdc++.h> #define maxn 10001 using namespace std; struct tree{ int data; struct tree*left,*right; }; int pre[maxn],in[maxn],n; void buildTree(tree**t,int preL,int preR,int inL,int inR) { if(preL>preR) return; int rootVal=pre[preL]; int rootIndex=inL; while(rootVal!=in[rootIndex]) rootIndex++; (*t)=(tree*)malloc(sizeof(tree)); (*t)->data=rootVal; (*t)->left=NULL; (*t)->right=NULL; buildTree(&(*t)->left,preL+1,preL+rootIndex-inL,inL,rootIndex-1); buildTree(&(*t)->right,preL+rootIndex-inL+1,preR,rootIndex+1,inR); } tree*findLCA(tree*t,int a,int b) { if(t==NULL) return NULL; if(t->data==a||t->data==b) return t; tree*left=findLCA(t->left,a,b); tree*right=findLCA(t->right,a,b); if(left&&right) return t; return left==NULL?right:left; } bool findNode(int x) { for(int i=0;i<n;i++) if(pre[i]==x) return true; return false; } int main() { int m,a,b; scanf("%d %d",&m,&n); for(int i=0;i<n;i++) scanf("%d",&in[i]); for(int i=0;i<n;i++) scanf("%d",&pre[i]); tree*root=NULL; buildTree(&root,0,n-1,0,n-1); while(m--) { scanf("%d %d",&a,&b); if(!findNode(a)&&!findNode(b)) printf("ERROR: %d and %d are not found.\n",a,b); else if(!findNode(a)||!findNode(b)) printf("ERROR: %d is not found.\n",findNode(a)==0?a:b); else { tree*lca=findLCA(root,a,b); if(lca->data==a||lca->data==b) printf("%d is an ancestor of %d.\n",lca->data,lca->data==a?b:a); else printf("LCA of %d and %d is %d.\n",a,b,lca->data); } } }