Leetcode: Lowest Common Ancestor of a Binary Search Tree

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.

如果如果p,q 比root小, 则LCA必定在左子树, 如果p,q比root大, 则LCA必定在右子树. 如果一大一小, 则root即为LCA. 如果p or q等于root，那么root也是LCA

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
12         if (root == null || p == null || q == null) return null;
13         if (root.val == p.val || root.val == q.val) return root;
14         else if (root.val > p.val && root.val > q.val) return lowestCommonAncestor(root.left, p, q);
15         else if (root.val < p.val && root.val < q.val) return lowestCommonAncestor(root.right, p, q);
16         else return root;
17     }
18 }

Iteration:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
12         if (root==null || p==null || q==null) return null;
13         while (root != null) {
14             if (root.val==p.val || root.val==q.val) return root;
15             if (root.val>p.val && root.val>q.val) root=root.left;
16             else if (root.val<p.val && root.val<q.val) root=root.right;
17             else return root;
18         }
19         return null;
20     }
21 }