235. Lowest Common Ancestor of a Binary Search Tree

一开始没发现是BST，以为是binary tree...草。。

public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        
        if(p.val < root.val && q.val < root.val) return lowestCommonAncestor(root.left,p,q);
        else if(p.val > root.val && q.val > root.val) return lowestCommonAncestor(root.right,p,q);
        else return root;
    }
}

---

---

---

二刷？其实不知道是几刷。

用的递归，就3种情况，左边，右边，当前。

/**
 * 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.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;
        }
        
    }
}

应该也可以迭代，试一下。
想多了，这不是BFS,DFS那种迭代递归的区别，这里迭代也很容易。。而且不需要额外的数据结构，省下了recursion里对于stack的空间消耗。

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

Time: iteratively O(n) recursively O(n) as well
Space: Iteratively better than recursively..