第6天 输入两棵二叉树A和B，判断B是不是A的子结构。

题目：输入两棵二叉树A和B，判断B是不是A的子结构。

代码中用到的二叉树源码：

http://blog.csdn.net/androiddevelop/article/details/18567085

/**
 *  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);
	}

}

参考资料：

《剑指offer》面试题18