剑指offer第二版-68.树中两个节点的最低公共祖先

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面试题68:树中两个节点的最低公共祖先

题目要求:
输入一棵树的根节点,输入两个被观察节点,求这两个节点的最低(最近)公共祖先。

解题思路:
此题比较开放,主要是对于“树”没有做明确说明,所以原书中就对树的可能情况做了假设,然后就衍生出多种思路

1)如果是二叉,搜索树:
    遍历找到比第一个节点大,比第二个节点小的节点即可
2)如果是父子间有双向指针的树:
    由下往上看,转化为找两个链表的第一个公共节点问题
3)如果只是一个包含父到子的指针的普通树:
   3.1)如果不能使用额外空间,从根节点开始判断他的子树是否包含那两个节点,找到最小的的子树即可
        时间复杂度o(n^2)(此为最差,平均不太好算。。。),空间复杂度为o(1)
   3.2) 如果能用额外空间,可以遍历两次(深度优先)获取根节点到那两个节点的路径,然后求两个路径的最后一个公共节点
        时间复杂度o(n),空间复杂度o(logn)

(1)(2)比较简单。下面仅对(3),以下图所示的树为例,进行思路实现与求解验证

                        A
                      /   \
                     B     C 
                  /     \
                D        E 
               / \     / | \
              F   G  H   I   J
package chapter7;

import java.util.*;

/**
 * Created with IntelliJ IDEA
 * Author: ryder
 * Date  : 2017/8/22
 * Time  : 17:15
 * Description:树中两个节点的最低公共祖先
 *
 **/
public class P326CommonParentInTree {
    public static class CommonTreeNode{
        public char val;
        public List<CommonTreeNode> children;
        public CommonTreeNode(char val){
            this.val = val;
            children = new LinkedList<>();
        }
        public void addChildren(CommonTreeNode... children){
            for(CommonTreeNode child:children)
                this.children.add(child);
        }
    }
    // 3.1所述的解法
    public static CommonTreeNode getLastParent1(CommonTreeNode root,CommonTreeNode node1,CommonTreeNode node2){
        if(root==null || node1==null || node2==null || !isInSubTree(root,node1,node2))
            return null;
        CommonTreeNode curNode = root;
        while (true){
            for(CommonTreeNode child:curNode.children){
                if(isInSubTree(child,node1,node2)){
                    curNode = child;
                    break;
                }
                if(child==curNode.children.get(curNode.children.size()-1))
                    return curNode;
            }
        }
    }
    public static boolean isInSubTree(CommonTreeNode root,CommonTreeNode node1,CommonTreeNode node2){
        Queue<CommonTreeNode> queue = new LinkedList<>();
        CommonTreeNode temp = null;
        int count = 0;
        queue.add(root);
        while (count!=2 && !queue.isEmpty()){
            temp = queue.poll();
            if(temp==node1||temp==node2)
                count++;
            if(!temp.children.isEmpty())
                queue.addAll(temp.children);
        }
        if(count==2)
            return true;
        return false;
    }
    // 3.2所述的解法
    public static CommonTreeNode getLastParent2(CommonTreeNode root,CommonTreeNode node1,CommonTreeNode node2){
        List<CommonTreeNode> path1 = new ArrayList<>();
        List<CommonTreeNode> path2 = new ArrayList<>();
        getPath(root,node1,path1);
        getPath(root,node2,path2);
        CommonTreeNode lastParent = null;
        for(int i=0;i<path1.size()&&i<path2.size();i++){
            if(path1.get(i)==path2.get(i))
                lastParent = path1.get(i);
            else
                break;
        }
        return lastParent;
    }
    public static boolean getPath(CommonTreeNode root,CommonTreeNode node,List<CommonTreeNode> curPath){
        if(root==node)
            return true;
        curPath.add(root);
        for(CommonTreeNode child:root.children){
            if(getPath(child,node,curPath))
                return true;
        }
        curPath.remove(curPath.size()-1);
        return false;
    }

    public static void main(String[] args){
        CommonTreeNode root = new CommonTreeNode('A');
        CommonTreeNode b = new CommonTreeNode('B');
        CommonTreeNode c = new CommonTreeNode('C');
        CommonTreeNode d = new CommonTreeNode('D');
        CommonTreeNode e = new CommonTreeNode('E');
        CommonTreeNode f = new CommonTreeNode('F');
        CommonTreeNode g = new CommonTreeNode('G');
        CommonTreeNode h = new CommonTreeNode('H');
        CommonTreeNode i = new CommonTreeNode('I');
        CommonTreeNode j = new CommonTreeNode('J');
        root.addChildren(b,c);
        b.addChildren(d,e);
        d.addChildren(f,g);
        e.addChildren(h,i,j);
        System.out.println(getLastParent1(root,f,h).val);
        System.out.println(getLastParent2(root,f,h).val);
        System.out.println(getLastParent1(root,h,i).val);
        System.out.println(getLastParent2(root,h,i).val);

    }
}

运行结果

B
B
E
E

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