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.
分析:
根据前序序列和中序序列从根节点开始不断地递归搜索左右子树地根节点，同时不断判断输入地两个孩子相对于当前根地位置。当当前根节点为其中之一或者两孩子分别位于当前根节点地左右子树地时候递归结束。
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 1e5 + 10; 4 int pre[maxn], in[maxn]; 5 int prenum[maxn], innum[maxn]; 6 map<int, int> vis; 7 int n, m; 8 int a, b; 9 10 int searchroot(int l, int r, int ll, int rr) 11 { 12 int rootpos = innum[pre[l]]; 13 int posa = innum[a], posb = innum[b]; 14 if ((rootpos - posa) * (rootpos - posb) <= 0) 15 return pre[l]; 16 if (rootpos > posa) 17 return searchroot(l + 1, l + rootpos - ll, ll, rootpos - 1); 18 else return searchroot(l + rootpos - ll + 1, r, rootpos + 1, rr); 19 } 20 21 int main() 22 { 23 cin >> m >> n; 24 for (int i = 1; i <= n; i++) 25 scanf("%d", in + i), vis[in[i]] = 1, innum[in[i]] = i; 26 for (int i = 1; i <= n; i++) 27 scanf("%d", pre + i), prenum[pre[i]] = i; 28 while (m--) 29 { 30 cin >> a >> b; 31 if (!vis[a] && !vis[b]) printf("ERROR: %d and %d are not found. ", a, b); 32 else if (!vis[a]) printf("ERROR: %d is not found. ", a); 33 else if (!vis[b]) printf("ERROR: %d is not found. ", b); 34 else 35 { 36 int res = searchroot(1, n, 1, n); 37 if (res == a) 38 printf("%d is an ancestor of %d. ", a, b); 39 else if (res == b) 40 printf("%d is an ancestor of %d. ", b, a); 41 else printf("LCA of %d and %d is %d. ", a, b, res); 42 } 43 } 44 }