经典算法- Lowest Common Ancestor of a Binary Search Tree（lct of bst）

题目描述：

一个二叉搜索树，给定两个节点a，b，求最小的公共祖先

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

例如：

2,8 ---->6 2,4----->2

原文描述：

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.

思路分析：

• 二叉树考虑递归的思路
• 如果a < = root <= b，可以确定root是lct
• 采用二分搜索的方式递归，root > a,b，继续遍历左子树，反之右边的子树
  -在递归的开始判断空值的情况，直接返回root

代码：

/**
 * 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 root;
        if(p.val > root.val && q.val > root.val)
            return lowestCommonAncestor(root.right, p, q);
        else if(p.val < root.val && q.val < root.val)
            return lowestCommonAncestor(root.left, p, q);
        else
            return root;
    }
}