Given a rooted tree with **N** nodes, the task is to find the **Lowest Common Ancestor** for a given set of nodes **V** of that tree.

**Examples:**

Input:1 / | \ 2 3 4 / \ | | 5 6 7 10 / \ 8 9 V[] = {7, 3, 8, 9}Output:3Input:1 / | \ 2 3 4 / \ | | 5 6 7 10 / \ 8 9 V[] = {4, 6, 7}Output:1

**Approach:** We can observe that

- the Lowest Common Ancestors for any set of nodes will have in-time less than or equal to that of all nodes in the set and out-time greater than or equal(equal if LCA is present in the set) to that of all nodes in the set.
- Thus, in order to solve the problem, we need to traverse the entire tree starting from the root node using Depth First Search Traversal and store the in-time, out-time, and level for every node.
- The node with in-time less than or equal to all the nodes in the set and out-time greater than or equal to all the nodes in the set and with maximum level possible is the answer.

Below is the implementation of the above approach:

## C++

`// C++ Program to find the LCA ` `// in a rooted tree for a given ` `// set of nodes ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Set time 1 initially ` `int` `T = 1; ` `void` `dfs(` `int` `node, ` `int` `parent, ` ` ` `vector<` `int` `> g[], ` ` ` `int` `level[], ` `int` `t_in[], ` ` ` `int` `t_out[]) ` `{ ` ` ` ` ` `// Case for root node ` ` ` `if` `(parent == -1) { ` ` ` `level[node] = 1; ` ` ` `} ` ` ` `else` `{ ` ` ` `level[node] = level[parent] + 1; ` ` ` `} ` ` ` ` ` `// In-time for node ` ` ` `t_in[node] = T; ` ` ` `for` `(` `auto` `i : g[node]) { ` ` ` `if` `(i != parent) { ` ` ` `T++; ` ` ` `dfs(i, node, g, ` ` ` `level, t_in, t_out); ` ` ` `} ` ` ` `} ` ` ` `T++; ` ` ` ` ` `// Out-time for the node ` ` ` `t_out[node] = T; ` `} ` ` ` `int` `findLCA(` `int` `n, vector<` `int` `> g[], ` ` ` `vector<` `int` `> v) ` `{ ` ` ` ` ` `// level[i]--> Level of node i ` ` ` `int` `level[n + 1]; ` ` ` ` ` `// t_in[i]--> In-time of node i ` ` ` `int` `t_in[n + 1]; ` ` ` ` ` `// t_out[i]--> Out-time of node i ` ` ` `int` `t_out[n + 1]; ` ` ` ` ` `// Fill the level, in-time ` ` ` `// and out-time of all nodes ` ` ` `dfs(1, -1, g, ` ` ` `level, t_in, t_out); ` ` ` ` ` `int` `mint = INT_MAX, maxt = INT_MIN; ` ` ` `int` `minv = -1, maxv = -1; ` ` ` `for` `(` `auto` `i = v.begin(); i != v.end(); i++) { ` ` ` `// To find minimum in-time among ` ` ` `// all nodes in LCA set ` ` ` `if` `(t_in[*i] < mint) { ` ` ` `mint = t_in[*i]; ` ` ` `minv = *i; ` ` ` `} ` ` ` `// To find maximum in-time among ` ` ` `// all nodes in LCA set ` ` ` `if` `(t_out[*i] > maxt) { ` ` ` `maxt = t_out[*i]; ` ` ` `maxv = *i; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Node with same minimum ` ` ` `// and maximum out time ` ` ` `// is LCA for the set ` ` ` `if` `(minv == maxv) { ` ` ` `return` `minv; ` ` ` `} ` ` ` ` ` `// Take the minimum level as level of LCA ` ` ` `int` `lev = min(level[minv], level[maxv]); ` ` ` `int` `node, l = INT_MIN; ` ` ` `for` `(` `int` `i = 1; i <= n; i++) { ` ` ` `// If i-th node is at a higher level ` ` ` `// than that of the minimum among ` ` ` `// the nodes of the given set ` ` ` `if` `(level[i] > lev) ` ` ` `continue` `; ` ` ` ` ` `// Compare in-time, out-time ` ` ` `// and level of i-th node ` ` ` `// to the respective extremes ` ` ` `// among all nodes of the given set ` ` ` `if` `(t_in[i] <= mint ` ` ` `&& t_out[i] >= maxt ` ` ` `&& level[i] > l) { ` ` ` `node = i; ` ` ` `l = level[i]; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `node; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `n = 10; ` ` ` `vector<` `int` `> g[n + 1]; ` ` ` `g[1].push_back(2); ` ` ` `g[2].push_back(1); ` ` ` `g[1].push_back(3); ` ` ` `g[3].push_back(1); ` ` ` `g[1].push_back(4); ` ` ` `g[4].push_back(1); ` ` ` `g[2].push_back(5); ` ` ` `g[5].push_back(2); ` ` ` `g[2].push_back(6); ` ` ` `g[6].push_back(2); ` ` ` `g[3].push_back(7); ` ` ` `g[7].push_back(3); ` ` ` `g[4].push_back(10); ` ` ` `g[10].push_back(4); ` ` ` `g[8].push_back(7); ` ` ` `g[7].push_back(8); ` ` ` `g[9].push_back(7); ` ` ` `g[7].push_back(9); ` ` ` ` ` `vector<` `int` `> v = { 7, 3, 8 }; ` ` ` ` ` `cout << findLCA(n, g, v) << endl; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

3

**Time Complexity:** *O(N)*

**Space Complexity:** *O(N)*

## Recommended Posts:

- Lowest Common Ancestor in a Binary Tree | Set 1
- Lowest Common Ancestor in a Binary Tree | Set 3 (Using RMQ)
- Lowest Common Ancestor in a Binary Search Tree.
- Lowest Common Ancestor in a Binary Tree | Set 2 (Using Parent Pointer)
- Least Common Ancestor of any number of nodes in Binary Tree
- Lowest Common Ancestor in Parent Array Representation
- Common nodes in the inorder sequence of a tree between given two nodes in O(1) space
- Tarjan's off-line lowest common ancestors algorithm
- Perform the given queries on the rooted tree
- Construct the Rooted tree by using start and finish time of its DFS traversal
- Print common nodes on path from root (or common ancestors)
- Minimum number of groups of nodes such that no ancestor is present in the same group
- Count the nodes of the tree which make a pangram when concatenated with the sub-tree nodes
- Kth ancestor of a node in binary tree | Set 2
- K-th ancestor of a node in Binary Tree | Set 3
- Construct tree from ancestor matrix
- K-th ancestor of a node in Binary Tree
- Query for ancestor-descendant relationship in a tree
- Construct Ancestor Matrix from a Given Binary Tree
- Maximum difference between node and its ancestor in Binary Tree

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.