Cracking the coding interview--Q4.7

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原文：

You have two very large binary trees: T1, with millions of nodes, and T2, with hundreds of nodes. Create an algorithm to decide if T2 is a subtree of T1

译文：

有两棵很大的二叉树：T1有上百万个结点，T2有上百个结点。写程序判断T2是否为T1的子树。

方法：

把T1，T2两颗树转换成字符串，如[head.data [左子树] [右子树]]，同时左右子树以相同的方法转换。

如果左子树或者右子树存在一颗并且只为一颗子树为空，则该子树设置为[null]，这样就能区分左子树或者右子树

然后判断T2是否为T1的子串即可。

package chapter_4_TreesandGraphs;import java.util.Scanner;/** *  * 有两棵很大的二叉树：T1有上百万个结点，T2有上百个结点。写程序判断T2是否为T1的子树。 *  */public class Question_4_7 {private static Node_4_6 t2Parent;public static void insertTreeNode(Node_4_6 curNode, Node_4_6 node) {if(node.data < curNode.data) {if(curNode.lchild != null) {insertTreeNode(curNode.lchild, node);} else{curNode.lchild = node;node.parent = curNode;}} else {if(curNode.rchild != null) {insertTreeNode(curNode.rchild, node);} else {curNode.rchild = node;node.parent = curNode;}}}public static void createTree(Node_4_6 head, int array[]) {int len = array.length;int curIndex = 1;while(curIndex < len) {Node_4_6 node = new Node_4_6();node.data = array[curIndex];if(node.data == 1) {t2Parent = node;System.out.println("set t2Parent.data = " + node.data);} node.lchild = null;node.rchild = null;insertTreeNode(head, node);curIndex ++;}}/** * @param node * @return *  * 把树转换成字符串 *  */public static String tranceString(Node_4_6 node) {if(node == null) {return "[null]";} else {if(node.lchild!=null || node.rchild!=null) {return "[" + node.data + tranceString(node.lchild) + tranceString(node.rchild) + "]";} else {return "[" + node.data + "]";}}}public static void main(String args[]) {Scanner scanner = new Scanner(System.in);String string = scanner.nextLine();// 7 10 1 3 6 5 4 2 8String strs[] = string.split(" ");int array[] = new int[strs.length];for(int i=0; i<array.length; i++) {array[i] = Integer.parseInt(strs[i]);}Node_4_6 head = new Node_4_6();head.data = array[0];head.lchild = null;head.rchild = null;createTree(head, array);String pattern1 = tranceString(head);String pattern2 = tranceString(t2Parent);System.out.println("T1 转换 ：" + pattern1);System.out.println("T2 转换 ：" + pattern2);System.out.println(pattern2 + "  是否为  " + pattern1 + " 子串");System.out.println(pattern1.contains(pattern2));}}

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