1151 LCA in a Binary Tree (30分)

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.

LCA算法，已知一个二叉树，进行查询两个节点是否有公共的祖先，有则返回，没有则声明一下。

#include <iostream>
using namespace std;
int M, N, in[10100], pre[10100], a, b;
struct node {
    int val;
    node *left, *right;
    node(int v): val(v), left(NULL), right(NULL){}
}*root;
node* build(int root, int start, int end) {
    if(start > end) return NULL;
    int i = start;
    while(i < end && in[i] != pre[root]) i++;
    node *n = new node(in[i]);
    n->left = build(root + 1, start, i - 1);
    n->right = build(root + 1 + (i - start), i + 1, end);
    return n;
}
bool find(node *root, int val) {
    if(!root) return false;
    if(root->val == val) return true;
    return find(root->left, val) || find(root->right, val);
}
node* lca(node *root, int a, int b) {
    if(!root) return NULL;
    if(root->val == a || root->val == b) return root;
    node *l = lca(root->left, a, b);
    node *r = lca(root->right, a, b);
    if(l && r) return root;
    return l == NULL ? r : l;
}
int main() {
    cin >> M >> N;
    for(int i = 0; i < N; i++)
        cin >> in[i];
    for(int i = 0; i < N; i++)
        cin >> pre[i];
    root = build(0, 0, N - 1);
    while(M--) {
        cin >> a >> b;
        bool right_a = find(root, a);
        bool right_b = find(root, b);
        if(right_a && right_b) {
            node *n = lca(root, a, b);
            if(n->val != a && n->val != b) printf("LCA of %d and %d is %d.
", a, b, n->val);
            else if(n->val == a) printf("%d is an ancestor of %d.
", a, b);
            else printf("%d is an ancestor of %d.
", b, a);
        } else {
            if(!right_a && !right_b) printf("ERROR: %d and %d are not found.
", a, b);
            else if(!right_a) printf("ERROR: %d is not found.
", a);
            else printf("ERROR: %d is not found.
", b);
        }
    }
    return 0;
}