Lowest Common Ancestor of a Binary Tree LeetCode Solution

Minimum Cost to Merge Stones

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


  • 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.

Lowest Common Ancestor of a Binary Tree Solutions

Time: O(h)
Space: O(h)


 class Solution {
  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;


 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;


  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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