1143. Lowest Common Ancestor (30)
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 (<= 1000), the number of pairs of nodes to be tested; and N (<= 10000), 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 99Sample 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.
给出一个搜索二叉树的前序遍历序列，给出两个数字，试图找出这两个数字在二叉树中的最近公共祖先（LCA）结点。
首先建立二叉树比较简单。然后通过findd函数去寻找最近公共祖先。findd函数有两个作用，f为1时是为了寻找a，找到了返回true，此时参数b用不到；f为0时是还在寻找两个点a和b的LCA，此时返回值用不到。当a和b一个比tree->data大另一个比tree->data小时，则当前节点的data就是a和b的LCA（如果a和b都在树中的话），所以接下来把f改为1继续调用findd函数来检查a和b是否都在二叉树中。
#include <cstdio>
#include <cstdlib>using namespace std;typedef struct node *NODE;
typedef struct node {int da;NODE l, r;
}node;NODE T;
int M, N;
int pre[10001];NODE buildtree(int s, int e) {if (e < s) return NULL;if (s == e) {NODE tnode = (NODE)malloc(sizeof(node));tnode->da = pre[s];tnode->l = NULL;tnode->r = NULL;return tnode;}int i = s + 1;while (i <= e && pre[i] < pre[s]) i++;NODE tnode = (NODE)malloc(sizeof(node));tnode->da = pre[s];tnode->l = buildtree(s + 1, i - 1);tnode->r = buildtree(i, e);return tnode;
}bool findd(NODE tree, int a, int b,int f) {if (f == 1) {if (tree == NULL) return false;if (tree->da == a) return true;return findd(a < tree->da ? tree->l : tree->r, a, 0, 1);}else {if (tree == NULL) {printf("ERROR: %d and %d are not found.\n", a, b);return true;}if (a < tree->da&&b < tree->da) findd(tree->l, a, b, 0);else if (a > tree->da&&b > tree->da) findd(tree->r, a, b, 0);else {bool fda = tree->da == a || findd(tree->l, a, 0, 1) || findd(tree->r, a, 0, 1);bool fdb = tree->da == b || findd(tree->l, b, 0, 1) || findd(tree->r, b, 0, 1);if (fda&&fdb) {if (tree->da == a || tree->da == b) printf("%d is an ancestor of %d.\n", tree->da == a ? a : b, tree->da == b ? a : b);elseprintf("LCA of %d and %d is %d.\n", a, b, tree->da);}else if (fda == false && fdb == false) {printf("ERROR: %d and %d are not found.\n", a, b);}else {printf("ERROR: %d is not found.\n", fda ? b : a);}return true;}}
}int main() {scanf("%d %d", &M, &N);int a, b;for (int i = 0; i < N; i++) {scanf("%d", &pre[i]);}T = buildtree(0, N - 1);for (int i = 0; i < M; i++) {scanf("%d %d", &a, &b);findd(T, a, b, 0);}return 0;
}