码迷,mamicode.com
首页 > 其他好文 > 详细

leetcode--Lowest Common Ancestor of a Binary Search Tree

时间:2015-08-01 13:02:22      阅读:98      评论:0      收藏:0      [点我收藏+]

标签:leetcode   数据结构   算法   

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.


题意:对于一棵BST树,给出树上的两个节点,找出它们的最接近的共同祖先节点。

分类:二叉树


解法1:由于题目给出的是BST树,根据BST树的性质,对于某个节点,其左子树上的所有节点的值都比它小,右子树上的所有节点的值都比它大。

对于根节点root而言,如果p,q两个相比root,一大一小,说明root就是它们的LCA

如果都在比root小,说明它们的LCA在root的左子树上。如果都比root大,说明它们的LCA在root的右子树上。

如果它们有其中一个跟root相等,说明这个就是LCA

根据这个思路,我们很快写出代码:

/**
 * 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) return null;
        if(p.val<root.val&&root.val<q.val){//如果一左一右
            return root;
        }else if(p.val>root.val&&root.val>q.val){//如果一左一右
            return root;
        }else if(p.val<root.val&&root.val>q.val){//如果都在左边,递归查找左子树
            return lowestCommonAncestor(root.left,p,q);
        }else if(p.val>root.val&&q.val>root.val){//如果都在右边,递归查找右子树
            return lowestCommonAncestor(root.right,p,q);
        }else if(p.val==root.val){//如果和root相等
            return p;
        }else if(q.val==root.val){//如果和root相等
            return q;
        }
        return null;
    }
}

解法2:解法2跟解法1的思路一样,但是精简了一下代码,除了在左右子树的情况,其余情况我们都可以返回root

/**
 * 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) return null;
        if(p.val<root.val&&root.val>q.val){//如果都在左边,递归查找左子树
            return lowestCommonAncestor(root.left,p,q);
        }else if(p.val>root.val&&q.val>root.val){//如果都在右边,递归查找右子树
            return lowestCommonAncestor(root.right,p,q);
        }else{//其余情况
            return root;
        }
    }
}


版权声明:本文为博主原创文章,未经博主允许不得转载。

leetcode--Lowest Common Ancestor of a Binary Search Tree

标签:leetcode   数据结构   算法   

原文地址:http://blog.csdn.net/crazy__chen/article/details/47184681

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!