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 a 2j
// / \ / \
// 2b 3c 4k 5l
// /\ / \
// 4d 5e 6f 7g
private 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 a 2j
// / \ / \
// 2b 3c 3k 5l
// /\ / \
// 4d 5e 6f 7g
private 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 7g
private 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 5l
private 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));
}

}