剑指offer(17)—树的子结构

树的子结构

题目描述

思路

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;
}
};