Well, remember to take advantage of the property of binary search trees, which is, node -> left -> val < node -> val < node -> right -> val
. Moreover, both p
and q
will be the descendants of the root
of the subtree that contains both of them. And the root
with the largest depth is just the lowest common ancestor. This idea can be turned into the following simple recursive code.
1 class Solution { 2 public: 3 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { 4 if (p -> val < root -> val && q -> val < root -> val) 5 return lowestCommonAncestor(root -> left, p, q); 6 if (p -> val > root -> val && q -> val > root -> val) 7 return lowestCommonAncestor(root -> right, p, q); 8 return root; 9 } 10 };
Of course, we may also solve it iteratively.
1 class Solution { 2 public: 3 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { 4 TreeNode* cur = root; 5 while (true) { 6 if (p -> val < cur -> val && q -> val < cur -> val) 7 cur = cur -> left; 8 else if (p -> val > cur -> val && q -> val > cur -> val) 9 cur = cur -> right; 10 else return cur; 11 } 12 } 13 };