标签：ber main nta int 第一个 esc test return order
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:
Given any two nodes in a BST, you are supposed to find their LCA.
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.
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..
6 8
6 3 1 2 5 4 8 7
2 5
8 7
1 9
12 -3
0 8
99 99
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:
给出一棵二叉查找树,判断所给出两个结点是不是在二叉树中,如果在则找出这两个结点的最近公共祖先节点.
思路:
构造二叉树的时候采用递归的方法,因为是先序遍历所以在先序遍历的序列中第一个大于父节点的结点的左侧为左子树,右侧为右子树.
寻找LCA的时候也是使用递归的方法,虽然上次做过一个相似的题目了,但是这次做的时候思路还是不太清楚,于是又看了一下之前的代码.总体的思路就是向子节点查找,如果结点的值和查找的值相等则返回这个节点,或者查找到最底层则返回NULL,否则就向下查找.返回的时候
return !left ? right : !right ? left : root;
意思就是,如果左子树为空,则返回右子树(不管其是不是空的),如果左子树非空的话,在查看右子树,如果右子树也非空的话则返回root,否则的话返回非空的左子树.
Code:
1 #include <iostream> 2 #include <vector> 3 4 using namespace std; 5 6 typedef struct Node* node; 7 struct Node { 8 int val; 9 node left; 10 node right; 11 Node(int v) { 12 val = v; 13 left = NULL; 14 right = NULL; 15 } 16 }; 17 18 node buildTree(vector<int>& v, int start, int end) { 19 if (start > end) return NULL; 20 node root = new Node(v[start]); 21 int pos = start + 1; 22 while (pos <= end && v[pos] < v[start]) ++pos; 23 if (pos == start + 1) { 24 root->right = buildTree(v, start + 1, end); 25 } else if (pos == end + 1) { 26 root->left = buildTree(v, start + 1, end); 27 } else { 28 root->left = buildTree(v, start + 1, pos - 1); 29 root->right = buildTree(v, pos, end); 30 } 31 return root; 32 } 33 34 bool foundNode(node root, int x) { 35 if (root == NULL) return false; 36 if (root->val == x) return true; 37 return foundNode(root->left, x) || foundNode(root->right, x); 38 } 39 40 node LCA(node root, int x, int y) { 41 if (root == NULL) return NULL; 42 if (root->val == x || root->val == y) return root; 43 node left = LCA(root->left, x, y); 44 node right = LCA(root->right, x, y); 45 return !left ? right : !right ? left : root; 46 } 47 48 int main() { 49 int M, N; 50 cin >> M >> N; 51 52 vector<int> num(N); 53 for (int i = 0; i < N; ++i) cin >> num[i]; 54 node root = buildTree(num, 0, N - 1); 55 // perOrder(root); 56 for (int i = 0; i < M; ++i) { 57 int x, y; 58 cin >> x >> y; 59 bool foundX = foundNode(root, x); 60 bool foundY = foundNode(root, y); 61 if (foundX && foundY) { 62 node lca = LCA(root, x, y); 63 if (lca->val == x) 64 cout << x << " is an ancestor of " << y << "." << endl; 65 else if (lca->val == y) 66 cout << y << " is an ancestor of " << x << "." << endl; 67 else 68 cout << "LCA of " << x << " and " << y << " is " << lca->val 69 << "." << endl; 70 } else if (foundX) { 71 cout << "ERROR: " << y << " is not found." << endl; 72 } else if (foundY) { 73 cout << "ERROR: " << x << " is not found." << endl; 74 } else { 75 cout << "ERROR: " << x << " and " << y << " are not found." << endl; 76 } 77 } 78 return 0; 79 }
标签：ber main nta int 第一个 esc test return order
原文地址：https://www.cnblogs.com/ruruozhenhao/p/12699922.html
