Day-05-二叉树与图 Leetcode-113, 236, 113, 199, 207

//预备知识:二叉树的构造
#include <stdio.h>
struct TreeNode {
       int val;
       TreeNode* left;
       TreeNode* right;
       TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
void preorder_print(TreeNode* node, int layer) {
       if (!node) {
              return;
       }
       for (int i = 0; i < layer; i++) {
              printf("------");//根据层数打印------
       }
       printf("[%d]\n", node->val);
       preorder_print(node->left, layer + 1);
       preorder_print(node->right, layer + 1);
}
void traversal(TreeNode* node) {
       if (!node) {
              return;
       }
       //前序遍历
       printf("[%d]\n", node->val);
       traversal(node->left);
       traversal(node->right);
       //     //中序遍历
       //     traversal(node->left);
       //     printf("[%d]\n", node->val);
       //     traversal(node->right);
       //
       //     //后序遍历
       //     traversal(node->left);
       //     traversal(node->right);
       //     printf("[%d]\n", node->val);
}
int main() {
       TreeNode a(1);
       TreeNode b(2);
       TreeNode c(5);
       TreeNode d(3);
       TreeNode e(4);
       TreeNode f(6);
       a.left = &b;
       a.right = &c;
       b.left = &d;
       b.right = &e;
       c.right = &f;
       preorder_print(&a, 0);
       return 0;
}

例一：LeetCode113

//给定一个二叉树与整数sum，找出所有从根节点到叶节点的路径，这些路径上累加值为sum
#include <stdio.h>
#include <vector>
struct TreeNode {
       int val;
       TreeNode* left;
       TreeNode* right;
       TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
       std::vector<std::vector<int>> pathSum(TreeNode* root, int sum) {
              std::vector<std::vector<int>> result;
              std::vector<int> path;
              int path_value = 0;
              preorder(root, path_value, sum, path, result);
              return result;
       }
private:
       void preorder(TreeNode* node, int& path_value, int sum, std::vector<int>& path, std::vector<std::vector<int>>& result) {
              if (!node) {
                     return;
              }
              path_value += node->val;
              path.push_back(node->val);
              if (!node->left && !node->right && path_value == sum) {
                     result.push_back(path);
              }
              preorder(node->left, path_value, sum, path, result);
              preorder(node->right, path_value, sum, path, result);
              path_value -= node->val;
              path.pop_back();
       }
};
int main() {
       TreeNode a(5);
       TreeNode b(4);
       TreeNode c(8);
       TreeNode d(11);
       TreeNode e(13);
       TreeNode f(4);
       TreeNode g(7);
       TreeNode h(2);
       TreeNode x(5);
       TreeNode y(1);
       a.left = &b;
       a.right = &c;
       b.left = &d;
       c.left = &e;
       c.right = &f;
       d.left = &g;
       d.left = &h;
       f.left = &x;
       f.right = &y;
       Solution solve;
       std::vector<std::vector<int>> result = solve.pathSum(&a, 22);
       for (int i = 0; i < result.size(); i++) {
              for (int j = 0; j < result[i].size(); j++) {
                     printf("[%d]", result[i][j]);
              }
              printf("\n");
       }
       return 0;
}

例二：LeetCode236

//已知二叉树，求二叉树中给定的两个节点的最近公共祖先
#include <stdio.h>
#include <vector>
struct TreeNode {
       int val;
       TreeNode* left;
       TreeNode* right;
       TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
       TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
              std::vector<TreeNode*> path;
              std::vector<TreeNode*> node_p_path;
              std::vector<TreeNode*> node_q_path;
              int finish = 0;
              preorder(root, p, path, node_p_path, finish);
              path.clear();
              finish = 0;
              preorder(root, q, path, node_q_path, finish);
              int path_len = 0;
              if (node_p_path.size() < node_q_path.size()) {
                     path_len = node_p_path.size();
              }
              else {
                     path_len = node_q_path.size();
              }
              TreeNode* result = 0;
              for (int i = 0; i < path_len; i++) {
                     if (node_p_path[i] == node_q_path[i]) {
                           result = node_p_path[i];
                     }
              }
              return result;
       }
private:
       void preorder(TreeNode* node, TreeNode* search, std::vector<TreeNode*>& path,
              std::vector<TreeNode*>& result, int& finish) {
              if (!node || finish) {
                     return;
              }
              path.push_back(node);
              if (node == search) {
                     finish = 1;
                     result = path;
              }
              preorder(node->left, search, path, result, finish);
              preorder(node->right, search, path, result, finish);
              path.pop_back();
       }
};
int main() {
       TreeNode a(3);
       TreeNode b(5);
       TreeNode c(1);
       TreeNode d(6);
       TreeNode e(2);
       TreeNode f(0);
       TreeNode x(8);
       TreeNode y(7);
       TreeNode z(4);
       a.left = &b;
       a.right = &c;
       b.left = &d;
       b.right = &e;
       c.left = &f;
       c.right = &x;
       e.left = &y;
       e.right = &z;
       Solution solve;
       TreeNode* result = solve.lowestCommonAncestor(&a, &b, &f);
       printf("lowestCommonAncestor = %d\n", result->val);
       result = solve.lowestCommonAncestor(&a, &d, &z);
       printf("lowestCommonAncestor = %d\n", result->val);
       result = solve.lowestCommonAncestor(&a, &b, &y);
       printf("lowestCommonAncestor = %d\n", result->val);
       return 0;
}

例三：LeetCode114

//给定一个二叉树，将该二叉树"就地"转化为单链表。
//单链表中节点序为前序遍历的顺序。
#include <stdio.h>
#include <vector>
struct TreeNode {
       int val;
       TreeNode* left;
       TreeNode* right;
       TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
       void flatten(TreeNode* root) {
              TreeNode* last = NULL;
              preorder(root, last);
       }
private:
       void preorder(TreeNode* node, TreeNode*& last) {
              if (!node) {
                     return;
              }
              if (node->left == NULL && node->right == NULL) {
                     last = node;
                     return;
              }
              TreeNode* left = node->left;
              TreeNode* right = node->right;
              TreeNode* left_last = NULL;
              TreeNode* right_last = NULL;
              if (left) {
                     preorder(left, left_last);
                     node->left = NULL;
                     node->right = left;
                     last = left_last;
              }
              if (right) {
                     preorder(right, right_last);
                     if (left_last) {
                           left_last->right = right;
                     }
                     last = right_last;
              }
       }
};
int main() {
       TreeNode a(1);
       TreeNode b(2);
       TreeNode c(5);
       TreeNode d(3);
       TreeNode e(4);
       TreeNode f(6);
       a.left = &b;
       a.right = &c;
       b.left = &d;
       b.right = &e;
       c.right = &f;
       Solution solve;
       solve.flatten(&a);
       TreeNode* head = &a;
       while (head) {
              if (head->left) {
                     printf("ERROR\n");
              }
              printf("[%d]", head->val);
              head = head->right;
       }
       printf("\n");
       return 0;
}

//二叉树的广度优先遍历，利用队列
#include <stdio.h>
#include <vector>
#include <queue>
struct TreeNode {
       int val;
       TreeNode* left;
       TreeNode* right;
       TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
void BFS_print(TreeNode* root) {
       std::queue<TreeNode*> Q;
       Q.push(root);
       while (!Q.empty()) {
              TreeNode* node = Q.front();
              Q.pop();
              printf("[%d]\n", node->val);
              if (node->left) {
                     Q.push(node->left);
              }
              if (node->right) {
                     Q.push(node->right);
              }
       }
}
int main() {
       TreeNode a(1);
       TreeNode b(2);
       TreeNode c(5);
       TreeNode d(3);
       TreeNode e(4);
       TreeNode f(6);
       a.left = &b;
       a.right = &c;
       b.left = &d;
       b.right = &e;
       c.right = &f;
       BFS_print(&a);
       return 0;
}

例四：LeetCode199

//给定一个二叉树，假设从该二叉树的右侧观察它
//将观察到的节点按照从上到下的顺序输出。
#include <stdio.h>
#include <vector>
#include <queue>
struct TreeNode {
       int val;
       TreeNode* left;
       TreeNode* right;
       TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};
class Solution {
public:
       std::vector<int> rightSideView(TreeNode* root) {
              std::vector<int> view;
              std::queue<std::pair<TreeNode*, int>> Q;
              if (root) {
                     Q.push(std::make_pair(root, 0));
              }
              while (!Q.empty()) {
                     TreeNode* node = Q.front().first;
                     int depth = Q.front().second;
                     Q.pop();
                     if (view.size() == depth) {
                           view.push_back(node->val);
                     }
                     else {
                           view[depth] = node->val;
                     }
                     if (node->left) {
                           Q.push(std::make_pair(node->left, depth + 1));
                     }
                     if (node->right) {
                           Q.push(std::make_pair(node->right, depth + 1));
                     }
              }
              return view;
       }
};
int main() {
       TreeNode a(1);
       TreeNode b(2);
       TreeNode c(5);
       TreeNode d(3);
       TreeNode e(4);
       TreeNode f(6);
       a.left = &b;
       a.right = &c;
       b.left = &d;
       b.right = &e;
       c.right = &f;
       Solution solve;
       std::vector<int> result = solve.rightSideView(&a);
       for (int i = 0; i < result.size(); i++) {
              printf("[%d]\n", result[i]);
       }
       return 0;
}

//图的邻接矩阵数据结构
#include <stdio.h>
#include <vector>
int main() {
       const int MAX_N = 5;
       int Graph[MAX_N][MAX_N] = { 0 };
       Graph[0][2] = 1;
       Graph[0][4] = 1;
       Graph[1][0] = 1;
       Graph[1][2] = 1;
       Graph[2][3] = 1;
       Graph[3][4] = 1;
       Graph[4][4] = 1;
       printf("Graph\n");
       for (int i = 0; i < MAX_N; i++) {
              for (int j = 0; j < MAX_N; j++) {
                     printf("%d ", Graph[i][j]);
              }
              printf("\n");
       }
       return 0;
}
//图的邻接表数据结构
struct GraphNode {
       int label;
       std::vector<GraphNode*> neighbors;
       GraphNode(int x) : label(x) {}
};
int main() {
       const int MAX_N = 5;
       GraphNode* Graph[MAX_N];
       for (int i = 0; i < MAX_N; i++) {
              Graph[i] = new GraphNode(i);
       }
       Graph[0]->neighbors.push_back(Graph[2]);
       Graph[0]->neighbors.push_back(Graph[4]);
       Graph[1]->neighbors.push_back(Graph[0]);
       Graph[1]->neighbors.push_back(Graph[2]);
       Graph[2]->neighbors.push_back(Graph[3]);
       Graph[3]->neighbors.push_back(Graph[4]);
       Graph[4]->neighbors.push_back(Graph[3]);
       printf("Graph:\n");
       for (int i = 0; i < MAX_N; i++) {
              printf("label[%d]:", i);
              for (int j = 0; j < Graph[i]->neighbors.size(); j++) {
                     printf(" %d", Graph[i]->neighbors[j]->label);
              }
              printf("\n");
       }
       for (int i = 0; i < MAX_N; i++) {
              delete Graph[i];
       }
       return 0;
}

//图的邻接矩阵数据结构
#include <stdio.h>
#include <vector>
#include <queue>
int main() {
       const int MAX_N = 5;
       int Graph[MAX_N][MAX_N] = { 0 };
       Graph[0][2] = 1;
       Graph[0][4] = 1;
       Graph[1][0] = 1;
       Graph[1][2] = 1;
       Graph[2][3] = 1;
       Graph[3][4] = 1;
       Graph[4][4] = 1;
       printf("Graph\n");
       for (int i = 0; i < MAX_N; i++) {
              for (int j = 0; j < MAX_N; j++) {
                     printf("%d ", Graph[i][j]);
              }
              printf("\n");
       }
       return 0;
}
图的邻接表数据结构
struct GraphNode {
       int label;
       std::vector<GraphNode*> neighbors;
       GraphNode(int x) : label(x) {}
};
int main() {
       const int MAX_N = 5;
       GraphNode* Graph[MAX_N];
       for (int i = 0; i < MAX_N; i++) {
              Graph[i] = new GraphNode(i);
       }
       Graph[0]->neighbors.push_back(Graph[2]);
       Graph[0]->neighbors.push_back(Graph[4]);
       Graph[1]->neighbors.push_back(Graph[0]);
       Graph[1]->neighbors.push_back(Graph[2]);
       Graph[2]->neighbors.push_back(Graph[3]);
       Graph[3]->neighbors.push_back(Graph[4]);
       Graph[4]->neighbors.push_back(Graph[3]);
       printf("Graph:\n");
       for (int i = 0; i < MAX_N; i++) {
              printf("label[%d]:", i);
              for (int j = 0; j < Graph[i]->neighbors.size(); j++) {
                     printf(" %d", Graph[i]->neighbors[j]->label);
              }
              printf("\n");
       }
       for (int i = 0; i < MAX_N; i++) {
              delete Graph[i];
       }
       return 0;
}
// DFS深度搜索邻接表
void DFS_graph(GraphNode* node, int visit[]) {
       visit[node->label] = 1;
       printf("%d ", node->label);
       for (int i = 0; i < node->neighbors.size(); i++) {
              if (visit[node->neighbors[i]->label] == 0) {
                     DFS_graph(node->neighbors[i], visit);
              }
       }
}
int main() {
       const int MAX_N = 5;
       GraphNode* Graph[MAX_N];
       for (int i = 0; i < MAX_N; i++) {
              Graph[i] = new GraphNode(i);
       }
       Graph[0]->neighbors.push_back(Graph[4]);
       Graph[0]->neighbors.push_back(Graph[2]);
       Graph[1]->neighbors.push_back(Graph[0]);
       Graph[1]->neighbors.push_back(Graph[2]);
       Graph[2]->neighbors.push_back(Graph[3]);
       Graph[3]->neighbors.push_back(Graph[4]);
       Graph[4]->neighbors.push_back(Graph[3]);
       int visit[MAX_N] = { 0 };
       for (int i = 0; i < MAX_N; i++) {
              if (visit[i] == 0) {
                     printf("From label(%d):", Graph[i]->label);
                     DFS_graph(Graph[i], visit);
                     printf("\n");
              }
       }
       for (int i = 0; i < MAX_N; i++) {
              delete Graph[i];
       }
       return 0;
}
// 图的广度优先遍历
void BFS_graph(GraphNode* node, int visit[]) {
       std::queue<GraphNode*> Q;
       Q.push(node);
       visit[node->label] = 1;
       while (!Q.empty()) {
              GraphNode* node = Q.front();
              Q.pop();
              printf("%d ", node->label);
              for (int i = 0; i < node->neighbors.size(); i++) {
                     if (visit[node->neighbors[i]->label] == 0) {
                           Q.push(node->neighbors[i]);
                           visit[node->neighbors[i]->label] = 1;
                     }
              }
       }
}
int main() {
       const int MAX_N = 5;
       GraphNode* Graph[MAX_N];
       for (int i = 0; i < MAX_N; i++) {
              Graph[i] = new GraphNode(i);
       }
       Graph[0]->neighbors.push_back(Graph[4]);
       Graph[0]->neighbors.push_back(Graph[2]);
       Graph[1]->neighbors.push_back(Graph[0]);
       Graph[1]->neighbors.push_back(Graph[2]);
       Graph[2]->neighbors.push_back(Graph[3]);
       Graph[3]->neighbors.push_back(Graph[4]);
       Graph[4]->neighbors.push_back(Graph[3]);
       int visit[MAX_N] = { 0 };
       for (int i = 0; i < MAX_N; i++) {
              if (visit[i] == 0) {
                     printf("From Label(%d):", Graph[i]->label);
                     BFS_graph(Graph[i], visit);
                     printf("\n");
              }
       }
       for (int i = 0; i < MAX_N; i++) {
              delete Graph[i];
       }
       return 0;
} 

例五：LeetCode207

//已知有n门课程，标记从0至n-1，课程之间是有依赖关系的，
//例如希望完成A课程，可能需要完成B课程。已知n个课程的依赖关系，
//可否将n门课程学完。
 
#include <stdio.h>
#include <vector>
 
struct GraphNode {
  int label;
  std::vector<GraphNode*> neighbors;
  GraphNode(int x) : label(x) {}
};
 
//深度优先算法
class Solution{
public:
  bool canFinish(int numCourses, std::vector<std::vector<int>> &prerequisites){
    std::vector<GraphNode*> graph;
    std::vector<int> visit;
    for(int i=0; i<numCourses; i++){
      graph.push_back(new GraphNode(i));
      visit.push_back(-1);
    }
    for(int i=0; i<prerequisites.size(); i++){
      GraphNode *begin = graph[prerequisites[i][1]];
      GraphNode *end = graph[prerequisites[i][0]];
      begin->neighbors.push_back(end);
    }
    for(int i=0; i<graph.size(); i++){
      if(visit[i] == -1 && !DFS_graph(graph[i], visit)){
        return false;
      }
    }
    for(int i=0; i<numCourses; i++){
      delete graph[i];
    }
    return true;
  }
 
private:
  bool DFS_graph(GraphNode* node, std::vector<int> &visit){
    visit[node->label] = 0;
    for(int i=0; i<node->neighbors.size(); i++){
      if(visit[node->neighbors[i]->label] == -1){
        if(DFS_graph(node->neighbors[i], visit) == 0){
          return false;
        }
      }
      else if(visit[node->neighbors[i]->label] == 0){
        return false;
      }
    }
    visit[node->label] = 1;
      return true;
    }
};

//广度搜索算法
class Solution{
public:
  bool canFinish(int numCourses, std::vector<std::vector<int>> &prerequisites){
    std::vector<GraphNode*> graph;
    std::vector<int> degree;
    for(int i=0; i<numCourses; i++){
      degree.push_back(0);
      graph.push_back(new GraphNode(i));
    }
    for(int i=0; i<prerequisites.size(); i++){
      GraphNode *begin = graph[prerequisites[i][1]];
      GraphNode *end = graph[prerequisites[i][0]];
      begin->neighbors.push_back(end);
      degree[prerequisites[i][0]]++;
    }
    std::queue<GraphNode*> Q;
    for(int i=0; i<numCourses; i++){
      if(degree[i] == 0){
        Q.push(graph[i]);
      }
    }
    while(!Q.empty()){
      GraphNode* node = Q.front();
      Q.pop();
      for(int i=0; i<node->neighbors.size(); i++){
        degree[node->neighbors[i]->label]--;
        if(degree[node->neighbors[i]->label] == 0){
          Q.push(node->neighbors[i]);
        }
      }
    }
    for(int i=0; i<graph.size(); i++){
      delete graph[i];
    }
    for(int i=0; i<degree.size(); i++){
      if(degree[i]){
        return false;
      }
    }
    return true;
  }
};

转载于:https://www.cnblogs.com/lihello/p/11520889.html