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

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

_______6______ / \ ___2__ ___8__ / \ / \ 0 _4 7 9 / \ 3 5

For example, the lowest common ancestor (LCA) of nodes

`2`

and `8`

is `6`

. Another example is LCA of nodes `2`

and `4`

is `2`

, since a node can be a descendant of itself according to the LCA definition.**Understand the problem:**

The problem can be solved by either treating the BST as a regular binary tree, or BST.

For the binary tree solution, we could check out the previous post using divide and conquer solution.

Now let's take a look at the solution for BST.

Code (Java):

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if (root == null) { return root; } if (root.val > p.val && root.val < q.val) { return root; } else if (root.val > p.val && root.val > q.val) { return lowestCommonAncestor(root.left, p, q); } else if (root.val < p.val && root.val < q.val) { return lowestCommonAncestor(root.right, p, q); } return root; } }

**Update on 1/27/16:**

I think this solution is clearer in the sense that we need to make sure node p and always at left of q.

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if (root == null || p == null || q == null) { return null; } // Make sure p is on the left of q if (p.val > q.val) { TreeNode temp = p; p = q; q = temp; } if (root.val >= p.val && root.val <= q.val) { return root; } else if (root.val > p.val && root.val > q.val) { return lowestCommonAncestor(root.left, p, q); } else { return lowestCommonAncestor(root.right, p, q); } } }

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