[面试题]Amazon: Given two binary trees，if the first tree is subtree of the second one

Given two binary trees, check if the first tree is subtree of the second one. A
subtree of a tree T is a tree S consisting of a node in T and all of its descendants in T. The subtree corresponding to the root node is the entire tree; the subtree corresponding to any other node is called a proper subtree.

两个树，先写另一个函数：isIdentical， 

用递归的方式：如何判断两个树是否完全一样呢？那就看从根开始，所有节点的值都相等。

写好了这个，就简单了。 给两个树的根，如何都相等，root就是root2的子树，反之，我们还要查看其他节点，要检查的子树root不变，看left和right树。

继续递归，只要找到一个相等的，那就是有子树。思路其实很简单，就是不要把这个问题放到一个递归里，比较麻烦。 我要是放一起考虑，就感觉会脑残。 

public static boolean isSubTree(TreeNode root,TreeNode root2){

if(root==null) return true;
if(root2==null) return false;

if(isIdentical(root,root2)) return true;
else {
return isIdentical(root,root2.left)||isIdentical(root,root2.right);
}

}
public static boolean isIdentical(TreeNode root,TreeNode root2){

if(root==null&&root2==null) return true;
if(root==null||root2==null) return false;

if(root.val==root2.val) {
return isSubTree(root.left,root2.left)&&isSubTree(root.right,root2.right);
}
else return false;

}

public class TreeNode {

int val;
TreeNode left;
TreeNode right;
TreeNode(int x){
val=x;
}

}

主函数test case：

public static void main(String[] args) {
// TODO Auto-generated method stub
TreeNode root=new TreeNode(10);
root.left=new TreeNode(4);
root.right=new TreeNode(6);
root.left.right=new TreeNode(30);

TreeNode root2=new TreeNode(26);
root2.left=new TreeNode(10);
root2.left.left=new TreeNode(4);
root2.left.right=new TreeNode(6);
root2.right=new TreeNode(6);
root2.left.left.right=new TreeNode(30);
root2.right=new TreeNode(3);
root2.right.right=new TreeNode(3);

System.out.println(isSubTree(root,root2));
//System.out.println(isIdentical(root,root2));
}