Description
You have two very large binary trees: T1, with millions of nodes, and T2, with hundreds of nodes. Create an algorithm to decide if T2 is a subtree of T1.
A tree T2 is a subtree of T1 if there exists a node n in T1 such that the subtree of n is identical to T2. That is, if you cut off the tree at node n, the two trees would be identical.
Example
T2 is a subtree of T1 in the following case:
1 3
/ \ /
T1 = 2 3 T2 = 4
/
4
T2 isn't a subtree of T1 in the following case:
1 3
/ \ \
T1 = 2 3 T2 = 4
/
4
解题：判断一个二叉树是不是另一个二叉树的问题。思路还是很清晰的，但是有些细节还是要注意下的。两个递归，一个用来判断两个树是否完全一样，这个只要不一样就返回false。
还有一个递归函数是用来返回最终结果的。区别在于，如果第二个函数判断不等，还有继续往T2的子树探索，知道找到一个与T1完全相等的，或者一直到最后也不存在这样的函数。
比较容易错的是，在判断结点值的时候，要先判断结点是不是null（判空）。
代码如下：
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/ public class Solution {
/**
* @param T1: The roots of binary tree T1.
* @param T2: The roots of binary tree T2.
* @return: True if T2 is a subtree of T1, or false.
*/
public boolean equal(TreeNode T1, TreeNode T2){
if(T1 == null && T2 == null){
return true;
}else if(T1 == null && T2 != null || T1 != null && T2 == null||T1.val != T2.val){
return false;
}else{
return equal(T1.left, T2.left) && equal(T1.right, T2.right);
}
}
public boolean isSubtree(TreeNode T1, TreeNode T2) {
// write your code here
if(T2 == null)
return true;
if(T1 == null)
return false;
if(equal(T1 , T2))
return true;
else return isSubtree(T1.left, T2)||isSubtree(T1.right, T2);
}
}