剑指offer(17)—树的子结构

树的子结构

题目描述

输入两棵二叉树A，B，判断B是不是A的子结构。（ps：我们约定空树不是任意一个树的子结构）

思路

递归

代码

/*struct TreeNode {    int val;    struct TreeNode *left;    struct TreeNode *right;    TreeNode(int x) :            val(x), left(NULL), right(NULL) {    }};*/class Solution {public:    bool DoesTree1HaveTree2(TreeNode* pRoot1, TreeNode* pRoot2){        if(pRoot2 == NULL)  //tree2是空树，不管tree1是否为空树，都返回true            return true;        if(pRoot1 == NULL)  //tree2不是空树，tree1是空树，则返回false            return false;        else{   //tree1和tree2都不是空树，则依次判断根节点，及左右子树是否一致            if(pRoot1->val == pRoot2->val)                return DoesTree1HaveTree2(pRoot1->left, pRoot2->left)                && DoesTree1HaveTree2(pRoot1->right, pRoot2->right);            else                return DoesTree1HaveTree2(pRoot1->left, pRoot2) || DoesTree1HaveTree2(pRoot1->right, pRoot2);        }    }    bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)    {        if(pRoot1 == NULL || pRoot2 == NULL)    //题意：空树不是任意一个树的子结构            return false;        bool res = false;        if(pRoot1->val == pRoot2->val)  //若根节点相同，则判断tree1是否包含tree2,调用函数DoesTree1HaveTree2            res = DoesTree1HaveTree2(pRoot1, pRoot2);        if(!res)    //若tree1不包含tree2，则判断tree2是否为tree1的左子树的子结构            res = HasSubtree(pRoot1->left, pRoot2);        if(!res)    //若tree2不是tree1的左子树的子结构，则判断tree2是否为tree1的右子树的子结构            res = HasSubtree(pRoot1->right, pRoot2);        return res;    }};