LeetCode OJ 235. 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.

由于这是一颗二叉搜索树，我们根据nodes v and w值的大小，能确定他们位于根节点的左边还是右边。如果两个节点都在根节点的左子树上，那么他们的LCA也肯定在根节点的左子树；如果两个节点都在根节点的右子树上，那么他们的LCA也肯定在根节点的右子树；如果两个节点分别位于根结的左右子树，那么根节点就是他们的LCA。否则的话直接返回根节点即可。代码如下：

/**
 * 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.val<root.val && q.val<root.val){       //在根节点左边
            return lowestCommonAncestor(root.left, p, q);
        }
        else if(p.val>root.val && q.val>root.val){  //在根节点右边
            return lowestCommonAncestor(root.right, p, q);
        }
        else return root;                           //在根节点两侧，或者其中一个节点与根节点相等
    }
}