第6天 输入两棵二叉树A和B，判断B是不是A的子结构。

/**
*  面试题18 输入两棵二叉树A和B，判断B是不是A的子结构。
*
*  2014-1-19
*/
public class SubstructureInTree<T extends Comparable<? super T>> {

public boolean hasSubtree(Node<T> root1, Node<T> root2) {

boolean result = false;

if (root1 != null && root2 != null) {

if (root1.getData().compareTo(root2.getData()) == 0) {
result = doesTree1HaveTree2(root1, root2);
//				System.out.println("value = " + result);
}
if (!result) {
result = hasSubtree(root1.getLeftNode(), root2);
}
if (!result) {
result = hasSubtree(root1.getRightNode(), root2);
}
}

return result;
}

private boolean doesTree1HaveTree2(Node<T> root1, Node<T> root2) {

if (root2 == null) {
return true;
}
if (root1 == null) {
return false;
}

if (root1.getData() != null && root2.getData() != null) {
if (root1.getData().compareTo(root2.getData()) != 0) {
return false;
}
}

return doesTree1HaveTree2(root1.getLeftNode(), root2.getLeftNode())
&& doesTree1HaveTree2(root1.getRightNode(), root2.getRightNode());
}

public static void main(String[] args) {
Integer[] data1 = {8, 8, 9, null, null, 2, 4, null, null, 7, null, null, 7, null, null};
BinaryTree<Integer> tree1 = new BinaryTree<Integer>();
tree1.createTree(data1);

System.out.println("树A前序遍历");
tree1.preorderTraversal(tree1.getRootNode());

Integer[] data2 = {2, 4, null, null, 7, null, null};
BinaryTree<Integer> tree2 = new BinaryTree<Integer>();
tree2.createTree(data2);

System.out.println("\n树B前序遍历");
tree2.preorderTraversal(tree2.getRootNode());

SubstructureInTree<Integer> inTree = new SubstructureInTree<Integer>();
boolean value = inTree.hasSubtree(tree1.getRootNode(), tree2.getRootNode());
System.out.println("\n" + value);
}

}