zoukankan      html  css  js  c++  java
  • **Lowest Common Ancestor of Two Nodes in a Binary Tree

    Given the root and two nodes in a Binary Tree. Find the lowest common ancestor(LCA) of the two nodes.

    The lowest common ancestor is the node with largest depth which is the ancestor of both nodes.

    Example

    For the following binary tree:

      4
     / 
    3   7
       / 
      5   6
    

    LCA(3, 5) = 4

    LCA(5, 6) = 7

    LCA(6, 7) = 7

    解法一: 

     1 /**
     2  * 本代码由九章算法编辑提供。没有版权欢迎转发。
     3  * - 九章算法致力于帮助更多中国人找到好的工作,教师团队均来自硅谷和国内的一线大公司在职工程师。
     4  * - 现有的面试培训课程包括:九章算法班,系统设计班,BAT国内班
     5  * - 更多详情请见官方网站:http://www.jiuzhang.com/
     6  */
     7 
     8 Version 1: Traditional Method
     9 
    10 public class Solution {
    11     private ArrayList<TreeNode> getPath2Root(TreeNode node) {
    12         ArrayList<TreeNode> list = new ArrayList<TreeNode>();
    13         while (node != null) {
    14             list.add(node);
    15             node = node.parent;
    16         }
    17         return list;
    18     }
    19     public TreeNode lowestCommonAncestor(TreeNode node1, TreeNode node2) {
    20         ArrayList<TreeNode> list1 = getPath2Root(node1);
    21         ArrayList<TreeNode> list2 = getPath2Root(node2);
    22         
    23         int i, j;
    24         for (i = list1.size() - 1, j = list2.size() - 1; i >= 0 && j >= 0; i--, j--) {
    25             if (list1.get(i) != list2.get(j)) {
    26                 return list1.get(i).parent;
    27             }
    28         }
    29         return list1.get(i+1);
    30     }
    31 }
    32 

     解法二:分治法

    public class Solution {
        /**
         * @param root: The root of the binary search tree.
         * @param A and B: two nodes in a Binary.
         * @return: Return the least common ancestor(LCA) of the two nodes.
         */
        public TreeNode lowestCommonAncestor(TreeNode root, TreeNode A, TreeNode B) {
            if (root == null){
                return null;
            }
            if (root == A || root == B){
                return root;//当找到一中一个node, 向上传递这个node
            }
            TreeNode left = lowestCommonAncestor(root.left, A, B);
            TreeNode right = lowestCommonAncestor(root.right, A, B);
            if (left != null && right != null){//check左右,如果左右分别包含两个node则这个root就是LCA, 向上传递这个node
                return root;
            } else if (left != null){//只有左边有node,向上传递此node
                return left;
            } else if (right != null){
                return right;
            } else{//左右边都没有node 传递null
                return null;
            }
        }
    }

     video: http://www.bittiger.io/videos/TLfAyc3Hxfaxp2Fs9/Jj33gh5tbGcFyKCWa

    reference:http://ryanleetcode.blogspot.com/2015/04/lowest-common-ancestor.html

  • 相关阅读:
    vue 按需加载
    需要打印真实尺寸大小等需求的,css的单位可以使用mm等做单位
    d3 比例尺
    d3 根据数据绘制svg
    d3 svg简单学习
    d3 使用随机数据生成条形图
    d3 画简单的柱形图
    d3 使用数据
    d3 数据绑定
    d3 添加元素相关api
  • 原文地址:https://www.cnblogs.com/hygeia/p/4772080.html
Copyright © 2011-2022 走看看