> For the complete documentation index, see [llms.txt](https://soumyajit4419.gitbook.io/ds-algo/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://soumyajit4419.gitbook.io/ds-algo/binary-tree/problems-related-to-binary-trees/lowest-common-ancestor-of-a-binary-tree.md).

# 2.Lowest Common Ancestor of a Binary Tree

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the [definition of LCA on Wikipedia](https://en.wikipedia.org/wiki/Lowest_common_ancestor): “The lowest common ancestor is defined between two nodes `p` and `q` as the lowest node in `T` that has both `p` and `q` as descendants (where we allow **a node to be a descendant of itself**).”

**Example 1:**

![](https://assets.leetcode.com/uploads/2018/12/14/binarytree.png)

```
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2018/12/14/binarytree.png)

```
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
```

**Example 3:**

```
Input: root = [1,2], p = 1, q = 2
Output: 1
```

**Lowest common ancestor** is defined between **two nodes p and q** as the **lowest node in T** that has **both p and q as descendants** (where we allow **a node to be a descendant of itself**).

## **Solution I**

**Method 1 (By Storing root to n1 and root to n2 paths):**\
**1)** Find path from root to n1 and store it in a vector or array.\
**2)** Find path from root to n2 and store it in another vector or array.\
**3)** Traverse both paths till the values in arrays are same. Return the common element just before the mismatch.

```cpp
class Solution
{
public:
    bool findPath(TreeNode *root, TreeNode *p, vector<TreeNode *> &v)
    {
        if (root == NULL)
        {
            return false;
        }

        v.push_back(root);

        if (root == p)
        {
            return true;
        }

      
        if (findPath(root->left, p, v))
        {
            return true;
        }
        
      
        if (findPath(root->right, p, v))
        {
            return true;
        }
        
        v.pop_back();
        return false;
    }

    TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *p, TreeNode *q)
    {
        vector<TreeNode *> v1;
        vector<TreeNode *> v2;
        TreeNode *res = NULL;

        findPath(root, p, v1);
        findPath(root, q, v2);
        
        int i;
        int x = min(v1.size(),v2.size());
        for (i = 0; i < x; i++)
        {
            if (v1[i] != v2[i])
            {
                break;
            }
        }
        res = v1[i - 1];
        return res;
    }
};
```

**Time Complexity O(n), and extra space of O(n)**\
**The tree is traversed twice, and then path arrays are compared.**

## Solution II

**Method 2 (Using Single Traversal)**

```cpp
class Solution
{
public:
    TreeNode *findLCA(TreeNode *root, TreeNode *p, TreeNode *q)
    {

        if (root == NULL)
        {
            return NULL;
        }

        if (root == p || root == q)
        {
            return root;
        }

        TreeNode *lt = findLCA(root->left, p, q);
        TreeNode *rt = findLCA(root->right, p, q);

        if (lt and rt)
        {
            return root;
        }

        if (lt)
        {
            return lt;
        }

        if (rt)
        {
            return rt;
        }

        return NULL;
    }

    TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *p, TreeNode *q)
    {

        TreeNode *t = findLCA(root, p, q);
        return t;
    }
};
```

**Time Complexity: O(n) , Const space used.**