# Lowest Common Ancestor of a Binary Tree LeetCode Solution Share:

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: “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:

```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:

```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
```

Constraints:

• The number of nodes in the tree is in the range `[2, 105]`.
• `-109 <= Node.val <= 109`
• All `Node.val` are unique.
• `p != q`
• `p` and `q` will exist in the tree.

Time: O(h)
Space: O(h)

### C++

`````` class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if (!root || root == p || root == q)
return root;

TreeNode* l = lowestCommonAncestor(root->left, p, q);
TreeNode* r = lowestCommonAncestor(root->right, p, q);

if (l && r)
return root;
return l ? l : r;
}
};
``````

### Java

`````` class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null || root == p || root == q)
return root;

TreeNode l = lowestCommonAncestor(root.left, p, q);
TreeNode r = lowestCommonAncestor(root.right, p, q);

if (l != null && r != null)
return root;
return l == null ? r : l;
}
}
``````

### Python

``````  class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
if not root or root == p or root == q:
return root

l = self.lowestCommonAncestor(root.left, p, q)
r = self.lowestCommonAncestor(root.right, p, q)

if l and r:
return root
return l or r
``````

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