LeetCode -- Lowest Common Ancestor of a Binary Search Tree

Question:

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。

Analysis:

问题描述：在一棵二叉搜索树中寻找两个节点的公共祖先（BST的特点是，左子树的所有节点的值小于根节点的值，右子树的所有节点的值大于根节点的值）。

1）先排除特殊情况；

2）如果根节点的值大于两个节点的最大值，则ancestor一定在左子树中，递归左子树；

3）如果根节点的值小于两个节点的最小值，则ancestor一定在右子树中，递归右子树；

4）否则，根节点与其中一个值相等或者大于左节点小于右节点，则结束递归，返回根节点。

Answer：

/**
 * 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(p == root || q == root) //如果一个是根节点，则根节点一定是最近common ancestor
        //         return root;
        if(root.val < Math.min(p.val, q.val)) //root比最小的都要小，说明在右边子树中
                return lowestCommonAncestor(root.right, p, q);
        else if(root.val > Math.max(p.val, q.val)) //root比最大的都要大，说明在左边子树中
                return lowestCommonAncestor(root.left, p, q);
        else return root;
    }
    
}

2)非递归方法判断（效率也没有很高啊。。）

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