LCA问题 详细
1、二叉搜索树上找两个节点LCA
1 public int query(Node t, Node u, Node v) {
2 int left = u.value;
3 int right = v.value;
4
5 //二叉查找树内，如果左结点大于右结点，不对，交换
6 if (left > right) {
7 int temp = left;
8 left = right;
9 right = temp;
10 }
11
12 while (true) {
13 //如果t小于u、v，往t的右子树中查找
14 if (t.value < left) {
15 t = t.right;
16
17 //如果t大于u、v，往t的左子树中查找
18 } else if (t.value > right) {
19 t = t.left;
20 } else {
21 return t.value;
22 }
23 }
24 }
2、二叉树上找两个节点
1 node* getLCA(node* root, node* node1, node* node2)
2 {
3 if(root == null)
4 return null;
5 if(root== node1 || root==node2)
6 return root;
7
8 node* left = getLCA(root->left, node1, node2);
9 node* right = getLCA(root->right, node1, node2);
10
11 if(left != null && right != null) // 两个点在root的左右两边，就是root了
12 return root;
13 else if(left != null) // 哪边不空返回哪边
14 return left;
15 else if (right != null)
16 return right;
17 else
18 return null;
19 }
朴素法求解： 先对树进行dfs标出每一个节点的深度，对于查找的两个点先判断在不在同一深度，不在 移到统一深度，然后在往上找
1 #include <iostream>
2 #include <cstdio>
3 #include <algorithm>
4 #include <cstdio>
5 #include <vector>
6 using namespace std;
7 const int Max = 10005;
8 int t, n, first, second, root;
9 vector<int> G[Max];
10 int indegree[Max], depth[Max], father[Max];
11 void inputTree()
12 {
13 for (int i = 0; i <= n; i++)
14 {
15 G[i].clear();
16 father[i] = 0;
17 indegree[i] = 0;
18 depth[i] = 0;
19 }
20 int u, v;
21 for (int i = 1; i < n; i++)
22 {
23 scanf("%d%d", &u, &v);
24 G[u].push_back(v);
25 indegree[v]++;
26 father[v] = u;
27 }
28 scanf("%d%d", &first, &second);
29 for (int i = 1; i <= n; i++)
30 {
31 if (indegree[i] == 0)
32 {
33 root = i;
34 break;
35 }
36 }
37 }
38 void dfs_depth(int u, int dep)
39 {
40 depth[u] = dep;
41 int Size = G[u].size();
42 for (int i = 0; i < Size; i++)
43 {
44 dfs_depth(G[u][i], dep + 1);
45 }
46 }
47 int find_ancestor()
48 {
49 while (depth[first] > depth[second])
50 {
51 first = father[first];
52 }
53 while (depth[first] < depth[second])
54 {
55 second = father[second];
56 }
57 while (first != second) // 这样直接返回first
58 {
59 first = father[first];
60 second = father[second];
61 }
62 return first;
63 }
64 int main()
65 {
66 scanf("%d", &t);
67 while (t--)
68 {
69 scanf("%d", &n);
70 inputTree();
71 dfs_depth(root, 0);
72 printf("%d\n", find_ancestor());
73 }
74 return 0;
75 }
tarjan + 并查集 解法：
1 #include <iostream>
2 #include <cstdio>
3 #include <algorithm>
4 #include <cstdio>
5 #include <vector>
6 using namespace std;
7 const int Max = 10005;
8 int t, n, first, second, root;
9 vector<int> G[Max], querry[Max];
10 int indegree[Max], father[Max], vis[Max];
11 void inputTree()
12 {
13 for (int i = 0; i <= n; i++)
14 {
15 G[i].clear();
16 querry[i].clear();
17 father[i] = i;
18 indegree[i] = 0;
19 vis[i] = 0;
20 }
21 int u, v;
22 for (int i = 1; i < n; i++)
23 {
24 scanf("%d%d", &u, &v);
25 G[u].push_back(v);
26 indegree[v]++;
27 }
28 scanf("%d%d", &first, &second);
29 querry[first].push_back(second);
30 querry[second].push_back(first);
31 for (int i = 1; i <= n; i++)
32 {
33 if (indegree[i] == 0)
34 {
35 root = i;
36 break;
37 }
38 }
39 }
40 int find_father(int x)
41 {
42 if (x == father[x])
43 return x;
44 return father[x] = find_father(father[x]);
45 }
46 void unionSet(int x, int y)
47 {
48 x = find_father(x);
49 y = find_father(y);
50 if (x != y)
51 father[y] = x;
52 }
53 void tarjan(int x)
54 {
55 int Size = G[x].size();
56 for (int i = 0; i < Size; i++)
57 {
58 int v = G[x][i];
59 tarjan(v);
60 unionSet(x, v);
61 }
62 vis[x] = 1;
63 /*
64 if (x == first && vis[second])
65 printf("%d\n", find_father(second));
66 else if (x == second && vis[first])
67 printf("%d\n", find_father(first));
68 */
69 Size = querry[x].size();
70 for (int i = 0; i < Size; i++)
71 {
72 if (vis[querry[x][i]])
73 {
74 printf("%d\n", find_father(querry[x][i]));
75 return;
76 }
77 }
78
79 }
80 int main()
81 {
82 scanf("%d", &t);
83 while (t--)
84 {
85 scanf("%d", &n);
86 inputTree();
87 tarjan(root);
88 }
89 return 0;
90 }
时间： 2024-10-19 10:53:14