题目1509：树中两个结点的最低公共祖先 剑指的40

题目1509：树中两个结点的最低公共祖先

时间限制：1 秒

内存限制：128 兆

特殊判题：否

提交：203

解决：57

题目描述：

给定一棵树，同时给出树中的两个结点，求它们的最低公共祖先。

输入：

输入可能包含多个测试样例。
对于每个测试案例，输入的第一行为一个数n(0<n<1000)，代表测试样例的个数。
其中每个测试样例包括两行，第一行为一个二叉树的先序遍历序列，其中左右子树若为空则用0代替，其中二叉树的结点个数node_num<10000。
第二行为树中的两个结点的值m1与m2(0<m1,m2<10000)。

输出：

对应每个测试案例，
输出给定的树中两个结点的最低公共祖先结点的值，若两个给定结点无最低公共祖先，则输出“My God”。

样例输入：

21 2 4 6 0 0 7 0 0 5 8 0 0 9 0 0 3 0 06 81 2 4 6 0 0 7 0 0 5 8 0 0 9 0 0 3 0 06 12

样例输出：

2My God

package Tree;import java.util.Iterator;import java.util.LinkedList;import java.util.Scanner;/** * @Title: Tree_findCommonAncestor.java * @Package Tree * @Description: Design an algorithm and write code to find the first common *               ancestor of two nodes in a binary tree Avoid storing additional *               nodes in a data structure NOTE: This is not necessarily a *               binary search tree * @author nutc * @date 2013-7-23 上午9:27:14 * @version V1.0 */public class Tree_findCommonAncestor2 {public static void main(String args[]){int n;Scanner sc = new Scanner(System.in);n = sc.nextInt();sc.nextLine(); //非常重要，读取多出来的回车！while(n>0){String[] str = sc.nextLine().split(" ");int [] num = new int[str.length];for(int i=0;i<num.length;i++){num[i]=Integer.parseInt(str[i]);}FatherNode root = buildTree(num);//root.display();int n1,n2 ;n1 = sc.nextInt();n2 = sc.nextInt();sc.nextLine();  //非常重要，读取多出来的回车！FatherNode com = findCom(root,n1,n2);if(com!=null)System.out.println(String.valueOf(com.value));else{System.out.println("My God");}n--;}}public static FatherNode buildTree(int[] num){if(num==null) return null;FatherNode root = new FatherNode(num[0]);FatherNode now = root;int index=1;while(index<num.length){//System.out.println(now.value);if(num[index]!=0){if(now.left==null){   //全部写反了。。。写成了 ！=now.left = new FatherNode(num[index]);now.left.father = now;  //父亲和儿子 不是可逆的！！！now = now.left;index ++;}else if(now.right==null){now.right = new FatherNode(num[index]);now.right.father = now;   //父亲和儿子 不是可逆的！！！now = now.right;index++;}else{now = now.father;}}else{now = now.father;index+=2;}}return root;}public static FatherNode findCom(FatherNode root,int n1,int n2){if(root==null) return null;LinkedList<FatherNode> list1 = new LinkedList<FatherNode>();LinkedList<FatherNode> list2 = new LinkedList<FatherNode>();findList(root,n1,list1);findList(root,n2,list2);//System.out.println(list1.toString());Iterator<FatherNode> it1 = list1.iterator();  //要限定类型啊亲！！！啊亲！！！Iterator<FatherNode> it2 = list2.iterator();FatherNode last = null;while(it1.hasNext() && it2.hasNext()){FatherNode f1 = it1.next();  //不是   it2 = it2.next();FatherNode f2 = it2.next();  //必须把node取出来比较，不能直接比较 it1 和 it2!!!//System.out.println(f1+" -- "+f2);if(f1 == f2){last = f1;}else{break;}}return last;}public static boolean findList(FatherNode root,int n,LinkedList<FatherNode> list){if(root ==null) return false;list.add(root);if(root.value == n){return true;}boolean iffound = false;iffound = findList(root.left,n,list);if(!iffound)iffound = findList(root.right,n,list);if(!iffound)  list.removeLast();  //remove 默认山第一个return iffound;}public static class FatherNode {FatherNode left;FatherNode right;FatherNode father;int value;public FatherNode(int v) {value = v;father = null;left = right = null;}public void display() {System.out.print("当前" + value);if (left != null) {System.out.print("  左边" + left.value);}if (right != null) {System.out.print("   右边" + right.value);}System.out.println();if (left != null)left.display();if (right != null)right.display();}public void addLeft(FatherNode n){left = n;n.father = this;}public void addRight(FatherNode n){right = n;n.father = this;}}}