1151 LCA in a Binary Tree (30point(s))

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：

根据先序遍历和中序遍历的结果，构建一棵二叉树，然后，在这颗二叉树中查找两个结点的最近公共祖先节点。

思路：

题目可以分成两部分组成

1. 先根据前序和中序构建一棵二叉树。

　　构建二叉树的时候采用递归的方式进行构建，根节点root在preorder中进行查找，再根据root在inorder中的位置确定左右子树中的节点个数。左子树的根节点就是其父节点root在preorder中的下标tag+1，右子树的根节点为tag + pos + 1，（pos为root在inorder中的下标）。递归跳出的条件是 start > end || tag >= inorder.size()。

2. 在在二叉树中查找两个结点的公共祖先节点。

　　https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-tree/comments/

Code：

#include<iostream>
#include<vector>
#include<set>

using namespace std;

typedef struct Node* node;

struct Node {
    int val;
    node left;
    node right;

    Node() {
        val = 0;
        left = NULL;
        right = NULL;
    }

    Node(int v) {
        val = v;
        left = NULL;
        right = NULL;
    }
};

vector<int> inorder, preorder;
int tag = 0;

node buildTree(int start, int end, int tag) {
    if (start >= end || tag >= inorder.size()) return NULL;
    int val = preorder[tag];
    node root = new Node(val);
    int lend, rstart, pos;
    for (int i = 0; i < inorder.size(); ++i) {
        if (inorder[i] == val) {
            pos = i;
            break;
        }
    }
    lend = pos - 1;
    rstart = pos + 1;
    root->left = buildTree(start, lend, tag+1);
    root->right = buildTree(rstart, end, tag+pos+1);
    return root;
}

node lowestCommonAncestor(node root, int n1, int n2) {
    if (!root || root->val == n1 || root->val == n2) return root;
    node left = lowestCommonAncestor(root->left, n1, n2);
    node right = lowestCommonAncestor(root->right, n1, n2);
    return !left ? right : !right ? left : root;
}

int main() {
    int m, n, t;
    cin >> m >> n;

    set<int> s;
    for (int i = 0; i < n; ++i) {
        cin >> t;
        inorder.push_back(t);
        s.insert(t);
    }
    for (int i = 0; i < n; ++i) {
        cin >> t;
        preorder.push_back(t);
    }

    node root = buildTree(0, n-1, 0);

    for (int i = 0; i < m; ++i) {
        int n1, n2;
        cin >> n1 >> n2;
        if (s.find(n1) != s.end() && s.find(n2) != s.end()) {
            node lca = lowestCommonAncestor(root, n1, n2);
            int v = lca->val;
            if (v == n1) {
                cout << n1 << " is an ancestor of " << n2 << "." << endl;
            } else if (v == n2) {
                cout << n2 << " is an ancestor of " << n1 << "." << endl;
            } else {
                cout << "LCA of " << n1 << " and " << n2 << " is " << v << "." << endl;
            }
        } else if (s.find(n1) != s.end()) {
            cout << "ERROR: " << n2 << " is not found." << endl;
        } else if (s.find(n2) != s.end()) {
            cout << "ERROR: " << n1 << " is not found." << endl;
        } else {
            cout << "ERROR: " << n1 << " and " << n2 << " are not found." << endl;
        }

    }

    return 0;
}

最后还是有一组数据没有通过。

---

建树的时候一定要注意小标的问题。

 1 #include <bits/stdc++.h>
 2
 3 using namespace std;
 4
 5 typedef struct Node* node;
 6
 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 vector<int> inOrder, preOrder;
19
20 node buildTree(int inl, int inr, int prel, int prer) {
21     // cout << prel << " " << prer << endl;
22     if (prel > prer || inl > inr) return NULL;
23     node root = new Node(preOrder[prel]);
24     int pos = 0;
25     for (int i = inl; i <= inr; ++i) {
26         if (inOrder[i] == preOrder[prel]) {
27             pos = i;
28             break;
29         }
30     }
31     int leftLen = pos - inl;
32     // cout << rightLen << " " << leftLen << endl;
33     root->left = buildTree(inl, pos - 1, prel + 1, prel + leftLen);
34     root->right = buildTree(pos + 1, inr, prel + leftLen + 1, prer);
35     return root;
36 }
37
38 node LCA(node root, int u, int v) {
39     if (!root || root->val == u || root->val == v) return root;
40     node left = LCA(root->left, u, v);
41     node right = LCA(root->right, u, v);
42     return !left ? right : !right ? left : root;
43 }
44
45 int main() {
46     int m, n;
47     cin >> m >> n;
48     inOrder.resize(n);
49     preOrder.resize(n);
50     for (int i = 0; i < n; ++i) cin >> inOrder[i];
51     for (int i = 0; i < n; ++i) cin >> preOrder[i];
52     set<int> visited(inOrder.begin(), inOrder.end());
53     node root = buildTree(0, n - 1, 0, n - 1);
54     int u, v;
55     for (int i = 0; i < m; ++i) {
56         cin >> u >> v;
57         if (visited.find(u) != visited.end() &&
58             visited.find(v) != visited.end()) {
59             node lca = LCA(root, u, v);
60             if (lca->val == u)
61                 cout << u << " is an ancestor of " << v << "." << endl;
62             else if (lca->val == v)
63                 cout << v << " is an ancestor of " << u << "." << endl;
64             else
65                 cout << "LCA of " << u << " and " << v << " is " << lca->val
66                      << "." << endl;
67         } else if (visited.find(u) != visited.end()) {
68             cout << "ERROR: " << v << " is not found." << endl;
69         } else if (visited.find(v) != visited.end()) {
70             cout << "ERROR: " << u << " is not found." << endl;
71         } else {
72             cout << "ERROR: " << u << " and " << v << " are not found." << endl;
73         }
74     }
75
76     return 0;
77 }

不用建树的代码：

 1 #include <iostream>
 2 #include <vector>
 3 #include <map>
 4 using namespace std;
 5 map<int, int> pos;
 6 vector<int> in, pre;
 7 void lca(int inl, int inr, int preRoot, int a, int b) {
 8     if (inl > inr) return;
 9     int inRoot = pos[pre[preRoot]], aIn = pos[a], bIn = pos[b];
10     if (aIn < inRoot && bIn < inRoot)
11         lca(inl, inRoot-1, preRoot+1, a, b);
12     else if ((aIn < inRoot && bIn > inRoot) || (aIn > inRoot && bIn < inRoot))
13         printf("LCA of %d and %d is %d.\n", a, b, in[inRoot]);
14     else if (aIn > inRoot && bIn > inRoot)
15         lca(inRoot+1, inr, preRoot+1+(inRoot-inl), a, b);
16     else if (aIn == inRoot)
17             printf("%d is an ancestor of %d.\n", a, b);
18     else if (bIn == inRoot)
19             printf("%d is an ancestor of %d.\n", b, a);
20 }
21 int main() {
22     int m, n, a, b;
23     scanf("%d %d", &m, &n);
24     in.resize(n + 1), pre.resize(n + 1);
25     for (int i = 1; i <= n; i++) {
26         scanf("%d", &in[i]);
27         pos[in[i]] = i;
28     }
29     for (int i = 1; i <= n; i++) scanf("%d", &pre[i]);
30     for (int i = 0; i < m; i++) {
31         scanf("%d %d", &a, &b);
32         if (pos[a] == 0 && pos[b] == 0)
33             printf("ERROR: %d and %d are not found.\n", a, b);
34         else if (pos[a] == 0 || pos[b] == 0)
35             printf("ERROR: %d is not found.\n", pos[a] == 0 ? a : b);
36         else
37             lca(1, n, 1, a, b);
38     }
39     return 0;
40 }