4.7-判断是否为子树

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.

参考：http://blog.youkuaiyun.com/fightforyourdream/article/details/20446739

其中有2个递归，isSubTree遍历递归来找到子树，这个过程需要每次判断当前t1中的当前子树是否和t2相同，也就是isSameTree递归。

时间复杂度O(mn)

#include <iostream>
#include <vector>
using namespace std;
struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
TreeNode *createTree(int *num, int left, int right)
{
    if (left > right)
        return NULL;

    int mid = (left + right) / 2;
    TreeNode *node = new TreeNode(num[mid]);
    node->left = createTree(num, left, mid - 1);
    node->right = createTree(num, mid + 1, right);
    return node;
}
TreeNode *sortedArrayToBST(int *num, int size)
{
    createTree(num, 0, size - 1);
}
//=================================
bool isSameTree(TreeNode *t1, TreeNode *t2)
{
    if(t1==NULL && t2==NULL)
        return true;
    if(t1==NULL || t2==NULL)
        return false;
    if(t1->val!=t2->val)
        return false;

    return isSameTree(t1->left, t2->left) && isSameTree(t1->right, t2->right);
}
bool isSubTree(TreeNode *t1, TreeNode *t2)
{
    if(t2==NULL)
        return true;
    if(t1==NULL)
        return false;
    if(t1->val==t2->val)
        if(isSameTree(t1, t2))
            return true;

    return isSubTree(t1->left, t2) || isSubTree(t1->right, t2);
}
int main()
{
    int num1[]= {1,2,3,4,5,6,7,8,9};
    TreeNode *root1 = sortedArrayToBST(num1,9);
    int num2[]= {6,7,8,9};
    TreeNode *root2 = sortedArrayToBST(num2,4);

    int num3[]= {3,4};
    TreeNode *root3 = sortedArrayToBST(num3,2);

    cout<<isSubTree(root1, root2)<<endl;
    cout<<isSubTree(root1, root3)<<endl;

    return 0;
}