剑指offer-面试题18-树的子结构

二叉树节点定义：

package case18_isTree1IsSubtreeOfTree2;public class myTreeNode {int data;myTreeNode lchild;myTreeNode rchild;public myTreeNode() {}public myTreeNode(int data) {this.data = data;}}

代码实现二叉树子树判断：

package case18_isTree1IsSubtreeOfTree2;public class Subtree {// (tree2,tree1) 树tree1是否是tree2的子树// tree2:.................tree1:// ..........8................8// ....... /...\ .......... /....\// .......8.....7..........9......2// ...../...\// ....9.....2// ........./..\// ........4....7/** * 在树tree2中寻找与树tree1的根值节点相同的节点。 *  * @param tree2被参考的大一点的树 * @param tree1子树 * @return true，tree1是tree2的子树 */private static boolean isTree1IsSubtreeOfTree2(myTreeNode tree2, myTreeNode tree1) {// 异常情况检测if (tree1 == null)return true;if (tree2 == null)return false;//boolean result = false;if (tree1.data == tree2.data)result = isDoesSubtree(tree2, tree1);if (!result)result = isTree1IsSubtreeOfTree2(tree2.lchild, tree1);if (!result)result = isTree1IsSubtreeOfTree2(tree2.rchild, tree1);return result;}/** * 判断以R为根节点的子树tree2和tree1是否具有相同的子结构。 *  * @param tree2,参考树的R节点 * @param tree1,子树的节点 * @return true,tree1是tree2的子树 */private static boolean isDoesSubtree(myTreeNode tree2, myTreeNode tree1) {// 这一步说明 tree1已经到达子节点if (tree1 == null)return true;// 这一步说明 tree2已经到达子节点，但是tree1没有到达子节点if (tree2 == null)return false;if (tree2.data != tree1.data)return false;// 递归判断tree1的左子节点，tree2的左子节点，他们是否相等boolean left = isDoesSubtree(tree2.lchild, tree1.lchild);// 递归判断tree1的右子节点，tree2的右子节点，他们是否相等boolean right = isDoesSubtree(tree2.rchild, tree1.rchild);return left && right;}public static void main(String[] args) {test1();test2();test3();test4();}//普通二叉树，树B是树A的子树//（数字是节点值，字母为节点名称）    //             1 a2j    //         /      \   /   \    //        2b      3c  4k   5l    //       /\       / \    //      4d 5e    6f  7gprivate static void test1(){myTreeNode a = new myTreeNode(1);myTreeNode b = new myTreeNode(2);myTreeNode c = new myTreeNode(3);myTreeNode d = new myTreeNode(4);myTreeNode e = new myTreeNode(5);myTreeNode f = new myTreeNode(6);myTreeNode g = new myTreeNode(7);a.lchild = b;a.rchild = c;b.lchild = d;b.rchild = e;e.lchild = f;e.rchild = g;myTreeNode j = new myTreeNode(2);myTreeNode k = new myTreeNode(4);myTreeNode l = new myTreeNode(5);j.lchild = k;j.rchild = l;System.out.print("//普通二叉树，树B是树A的子树--test1():");System.out.println(isTree1IsSubtreeOfTree2(a, j));}//普通二叉树，树B 不 是树A的子树//（数字是节点值，字母为节点名称）//树A：树B    //             1 a2j    //         /      \   /   \    //        2b      3c  3k   5l    //       /\       / \    //      4d 5e    6f  7gprivate static void test2(){myTreeNode a = new myTreeNode(1);myTreeNode b = new myTreeNode(2);myTreeNode c = new myTreeNode(3);myTreeNode d = new myTreeNode(4);myTreeNode e = new myTreeNode(5);myTreeNode f = new myTreeNode(6);myTreeNode g = new myTreeNode(7);a.lchild = b;a.rchild = c;b.lchild = d;b.rchild = e;e.lchild = f;e.rchild = g;myTreeNode j = new myTreeNode(2);myTreeNode k = new myTreeNode(3);myTreeNode l = new myTreeNode(5);j.lchild = k;j.rchild = l;System.out.print("普通二叉树，树B 不 是树A的子树--test2():");System.out.println(isTree1IsSubtreeOfTree2(a, j));}//普通二叉树，树B是普通二叉树，树A是null//（数字是节点值，字母为节点名称）//树A：树B：为null    //             1 a    //         /      \    //        2b      3c    //       /\       / \    //      4d 5e    6f  7gprivate static void test3(){myTreeNode j = new myTreeNode(2);myTreeNode k = new myTreeNode(4);myTreeNode l = new myTreeNode(5);j.lchild = k;j.rchild = l;System.out.print("树B是普通二叉树，树A是null--test3():");System.out.println(isTree1IsSubtreeOfTree2(null,j));}//普通二叉树，树B null，树A是普通二叉树//（数字是节点值，字母为节点名称）//树A：null, 树B:    //2j    //   /   \    //  4k   5lprivate static void test4(){myTreeNode a = new myTreeNode(1);myTreeNode b = new myTreeNode(2);myTreeNode c = new myTreeNode(3);myTreeNode d = new myTreeNode(4);myTreeNode e = new myTreeNode(5);myTreeNode f = new myTreeNode(6);myTreeNode g = new myTreeNode(7);a.lchild = b;a.rchild = c;b.lchild = d;b.rchild = e;e.lchild = f;e.rchild = g;System.out.print("树B null，树A是普通二叉树--test4():");System.out.println(isTree1IsSubtreeOfTree2(a,null));}}