zoukankan      html  css  js  c++  java
  • Leetcode: 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.

    如果如果p,q 比root小, 则LCA必定在左子树, 如果p,q比root大, 则LCA必定在右子树. 如果一大一小, 则root即为LCA. 如果p or q等于root,那么root也是LCA

     1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     int val;
     5  *     TreeNode left;
     6  *     TreeNode right;
     7  *     TreeNode(int x) { val = x; }
     8  * }
     9  */
    10 public class Solution {
    11     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    12         if (root == null || p == null || q == null) return null;
    13         if (root.val == p.val || root.val == q.val) return root;
    14         else if (root.val > p.val && root.val > q.val) return lowestCommonAncestor(root.left, p, q);
    15         else if (root.val < p.val && root.val < q.val) return lowestCommonAncestor(root.right, p, q);
    16         else return root;
    17     }
    18 }

     Iteration:

     1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     int val;
     5  *     TreeNode left;
     6  *     TreeNode right;
     7  *     TreeNode(int x) { val = x; }
     8  * }
     9  */
    10 public class Solution {
    11     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    12         if (root==null || p==null || q==null) return null;
    13         while (root != null) {
    14             if (root.val==p.val || root.val==q.val) return root;
    15             if (root.val>p.val && root.val>q.val) root=root.left;
    16             else if (root.val<p.val && root.val<q.val) root=root.right;
    17             else return root;
    18         }
    19         return null;
    20     }
    21 }
  • 相关阅读:
    C语言中的排序算法--冒泡排序,选择排序,希尔排序
    常见算法:C语言求最小公倍数和最大公约数三种算法
    提高软件测试效率的方法探讨
    面试官询问的刁钻问题——以及如何巧妙地应付它们
    软件测试面试--如何测试网页的登录页面
    如何衡量测试效率,如何提高测试效率!
    利用交叉测试提升软件测试效率
    交叉测试的必要性和遇到的问题
    敏捷测试
    HttpWatch工具简介及使用技巧
  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/5060239.html
Copyright © 2011-2022 走看看