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

/**
 * 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;
        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);
        else return root;
    }
}

 Iteration:

/**
 * 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;
        while (root != null) {
            if (root.val==p.val || root.val==q.val) return root;
            if (root.val>p.val && root.val>q.val) root=root.left;
            else if (root.val<p.val && root.val<q.val) root=root.right;
            else return root;
        }
        return null;
    }
}