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
andq
will exist in the tree.
Lowest Common Ancestor of a Binary Tree Solutions
✅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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