剑指offer第二版-26树的子结构

/**
 * 树的子结构
 * <p>
 * 输入两棵二叉树A和B，判断B是不是A的子结构。
 * <p>
 * 当A有一个节点与B的根节点值相同时，则需要从A的那个节点开始严格匹配，对应于下面的tree1HasTree2FromRoot函数。
 * 如果匹配不成功，则返回到开始匹配的那个节点，对它的左右子树继续判断是否与B的根节点值相同，重复上述过程。
 * 应注意，必须先比较root2的节点是否为空
 *
 * @author VicterTian
 * @version V1.0
 * @Date 2019/2/26
 */
public class P148_SubstructureInTree {
  static class TreeNode<T> {
    T val;
    TreeNode<T> left;
    TreeNode<T> right;

    TreeNode(T val) {
      this.left = null;
      this.right = null;
      this.val = val;
    }
  }

  private static boolean hasSubtree(TreeNode<Integer> root1, TreeNode<Integer> root2) {
    if (root2 == null) {
      return true;
    }
    if (root1 == null) {
      return false;
    }
    if (root1.val.equals(root2.val)) {
      if (tree1HasTree2FromRoot(root1, root2)) {
        return true;
      }
    }
    return hasSubtree(root1.left, root2) || hasSubtree(root1.right, root2);
  }

  private static boolean tree1HasTree2FromRoot(TreeNode<Integer> root1, TreeNode<Integer> root2) {
    if (root2 == null) {
      return true;
    }
    if (root1 == null) {
      return false;
    }

    return root1.val.equals(root2.val) && tree1HasTree2FromRoot(root1.left, root2.left) && tree1HasTree2FromRoot(root1.right, root2.right);
  }

  public static void main(String[] args) {
    TreeNode<Integer> root1 = new TreeNode<>(8);
    root1.left = new TreeNode<>(8);
    root1.right = new TreeNode<>(7);
    root1.left.left = new TreeNode<>(9);
    root1.left.right = new TreeNode<>(2);
    root1.left.right.left = new TreeNode<>(4);
    root1.left.right.right = new TreeNode<>(7);
    TreeNode<Integer> root2 = new TreeNode<>(8);
    root2.left = new TreeNode<>(9);
    root2.right = new TreeNode<>(2);
    TreeNode<Integer> root3 = new TreeNode<>(2);
    root3.left = new TreeNode<>(4);
    root3.right = new TreeNode<>(3);
    System.out.println(hasSubtree(root1, root2));
    System.out.println(hasSubtree(root1, root3));
  }
}