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本帖最后由 糖逗 于 2020-5-8 17:29 编辑
题目描述:
- 输入两棵二叉树A和B,判断B是不是A的子结构。(约定空树不是任意一个树的子结构)
- B是A的子结构, 即 A中有出现和B相同的结构和节点值。
- 例如:
- 给定的树 A:
-      3
-     / \
-    4   5
-   / \
-  1   2
- 给定的树 B:
-    4 
-   /
-  1
- 返回 true,因为 B 与 A 的一个子树拥有相同的结构和节点值。
- 示例 1:
- 输入:A = [1,2,3], B = [3,1]
- 输出:false
- 示例 2:
- 输入:A = [3,4,5,1,2], B = [4,1]
- 输出:true
- 限制:
- 0 <= 节点个数 <= 10000
- 来源:力扣(LeetCode)
- 链接:https://leetcode-cn.com/problems/shu-de-zi-jie-gou-lcof
- 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
复制代码
- #include<iostream>
- #include <malloc.h>
- #include <vector>
- #include <math.h>
- #include <queue>
- using namespace std;
- struct TreeNode{
- int val;
- TreeNode* left;
- TreeNode* right;
- TreeNode(int x): val(x), left(NULL), right(NULL){
- }
- };
- TreeNode* CreateTree(vector<int> input){
- TreeNode* res = (TreeNode*)malloc(sizeof(TreeNode)*input.size());
- for(int i = 0; i < input.size(); i++){
- res[i].val = input[i];
- res[i].left = NULL;
- res[i].right = NULL;
- }
- for(int i= 0; i < input.size(); i++){
- if(2*i+1 < input.size()){
- res[i].left = &res[2*i+1];
- }
- if(2*i+2 < input.size()){
- res[i].right = &res[2*i+2];
- }
-
- }
- return res;
-
- }
- void middle(TreeNode* root, vector<vector<int> >& res, int left, int right, int depth){
- if(root == NULL) return;
- int insert = left + (right - left) / 2;
- res[depth][insert] = root->val;
-
- middle(root->left, res, left, insert - 1, depth + 1);
- middle(root->right, res, insert + 1, right, depth + 1);
- }
- int treeDepth(TreeNode* root){
- if(root == NULL || root -> val == 0) return 0;
- return max(treeDepth(root->left) + 1, treeDepth(root->right) + 1);
- }
-
- void PrintTree(TreeNode* root) {
- int depth = treeDepth(root);
- int width = pow(2, depth) - 1;
- vector<vector<int> > res(depth, vector<int>(width, 0));
- middle(root, res, 0, width - 1, 0);
- for(int i = 0; i < res.size(); i++){
- for(int j = 0; j < res[i].size();j++){
- if(res[i][j] == 0){
- cout << " ";
- }
- else{
- cout << res[i][j];
- }
-
- }
- cout << endl;
- }
- cout << "------------------" << endl;
- }
- TreeNode* findNode(TreeNode* A, TreeNode* B){
- if(A == NULL) return A;
- queue <TreeNode*> temp;
- temp.push(A);
- while(!temp.empty()){
- auto node = temp.front();
- temp.pop();
- if(node -> val == B -> val) return node;
- if(node ->left != NULL) temp.push(node->left);
- if(node -> right != NULL) temp.push(node -> right);
- }
- return NULL;
- }
- bool solution(TreeNode*A, TreeNode*B){
- if(A == NULL&& B!= NULL) return false;
- if(A != NULL && B == NULL) return true;
- if(A==NULL && B ==NULL) return true;
- if(A->val != B->val) return false;
- return solution(A->right,B->right) && solution(A->left, B->left);
- }
- bool isSubStructure(TreeNode*A, TreeNode*B){
- if(A!=NULL && B==NULL) return false;
- TreeNode* node = findNode(A, B);
- return solution(node, B);
-
- }
- int main(void){
- vector<int> input1, input2;
- cout << "send numbers for the first tree" << endl;
- int number1;
- while(cin >> number1){
- input1.push_back(number1);
- }
- cin.clear();
- cout << "send numbers for the second tree" << endl;
- int number2;
- while(cin >> number2){
- input2.push_back(number2);
- }
-
- TreeNode* root1 = CreateTree(input1);
- PrintTree(root1);
-
- TreeNode* root2 = CreateTree(input2);
- PrintTree(root2);
-
- bool res = isSubStructure(root1, root2);
- cout << res << endl;
- return 0;
- }
复制代码
注意事项:
1.解题思路:分两步,先找到A中包含B根节点的子节点,然后再递归比较A的子树C和B,判断B是否是C的子树。
2.第一步找子节点用到广度优先搜索。
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