子树判断

有两个不同大小的二进制树： T1 有上百万的节点； T2有好几百的节点。请设计一种算法，判定 T2 是否为 T1的子树。

public class Solution {
    /**
     * @param T1, T2: The roots of binary tree.
     * @return: True if T2 is a subtree of T1, or false.
     */
    public boolean doesT1ContainsT2(TreeNode T1,TreeNode T2){
        if(T2 == null && T1 == null) return true;
        if(T1 != null && T2 != null && T1.val == T2.val) {
            return doesT1ContainsT2(T1.left,T2.left) && doesT1ContainsT2(T1.right,T2.right);
        }
        return false;
    }
    public boolean isSubtree(TreeNode T1, TreeNode T2) {
        // write your code here
        if(T2 == null){
            return true;
        }else if(T1 == null){
            return false;
        }
        boolean result = false;
        if(T1.val == T2.val){
            result = doesT1ContainsT2(T1,T2);
        }
        if(!result){
            result = isSubtree(T1.left,T2);
        }
        if(!result){
            result = isSubtree(T1.right,T2);
        }
        return result;
    }
}