《Cracking the Coding Interview》——第4章：树和图——题目9

2014-03-19 05:07

题目：给定一棵二叉树T和一个值value，在T中找出所有加起来和等于value的路径。路径的起点和终点都可以是树的任意节点。

解法：我偷了个懒，直接把这棵树看成一个无向图，用DFS来进行暴力搜索解决问题。因为没有什么数据顺序或是范围的限制，所以搜索剪枝好像也不太容易。

代码：

 // 4.9 Find all paths in a binary tree, the path doesn't have to start or end at the root or a leaf node.
 #include <cstdio>
 #include <unordered_map>
 #include <unordered_set>
 using namespace std;

 struct TreeNode {
     int val;
     TreeNode *left;
     TreeNode *right;

     TreeNode(int _val = ):val(_val), left(nullptr), right(nullptr) {};
 };

 void constructTree(TreeNode *&root)
 {
     int val;

     scanf("%d", &val);
     if (val == ) {
         root = nullptr;
     } else {
         root = new TreeNode(val);
         constructTree(root->left);
         constructTree(root->right);
     }
 }

 void constructGraph(TreeNode *root, unordered_map<TreeNode *, unordered_set<TreeNode *> > &graph)
 {
     if (root->left != nullptr) {
         graph[root].insert(root->left);
         graph[root->left].insert(root);
         constructGraph(root->left, graph);
     }

     if (root->right != nullptr) {
         graph[root].insert(root->right);
         graph[root->right].insert(root);
         constructGraph(root->right, graph);
     }
 }

 void DFS(TreeNode *node, vector<TreeNode *> &path, int sum, const int target,
          unordered_set<TreeNode *> &checked,
          unordered_map<TreeNode *, unordered_set<TreeNode *> > &graph, vector<vector<TreeNode *> > &result)
 {
     path.push_back(node);
     checked.insert(node);

     if (sum == target) {
         result.push_back(path);
     }
     unordered_set<TreeNode *>::iterator it;
     for (it = graph[node].begin(); it != graph[node].end(); ++it) {
         if (checked.find(*it) == checked.end()) {
             DFS(*it, path, sum + (*it)->val, target, checked, graph, result);
         }
     }

     checked.erase(node);
     path.pop_back();
 }

 void doDFS(TreeNode *root, vector<TreeNode *> &path, const int target,
          unordered_set<TreeNode *> &checked,
          unordered_map<TreeNode *, unordered_set<TreeNode *> > &graph, vector<vector<TreeNode *> > &result)
 {
     path.clear();
     checked.clear();
     DFS(root, path, root->val, target, checked, graph, result);
     if (root->left != nullptr) {
         doDFS(root->left, path, target, checked, graph, result);
     }
     if (root->right != nullptr) {
         doDFS(root->right, path, target, checked, graph, result);
     }
 }

 void clearTree(TreeNode *&root)
 {
     if (root == nullptr) {
         return;
     }

     if (root->left != nullptr) {
         clearTree(root->left);
     }
     if (root->right != nullptr) {
         clearTree(root->right);
     }
     delete root;
     root = nullptr;
 }

 int main()
 {
     int i, j;
     int target;
     TreeNode *root;
     unordered_map<TreeNode *, unordered_set<TreeNode *> > graph;
     unordered_map<TreeNode *, unordered_set<TreeNode *> >::iterator it;
     unordered_set<TreeNode *> checked;
     vector<TreeNode *> path;
     vector<vector<TreeNode *> > result;

     while (true) {
         constructTree(root);
         if (root == nullptr) {
             break;
         }
         constructGraph(root, graph);
         while (scanf("%d", &target) ==  && target) {
             doDFS(root, path, target, checked, graph, result);
             if (result.empty()) {
                 printf("No path is found.\n");
             } else {
                 for (i = ; i < (int)result.size(); ++i) {
                     printf("%d", result[i][]->val);
                     for (j = ; j < (int)result[i].size(); ++j) {
                         printf("->%d", result[i][j]->val);
                     }
                     result[i].clear();
                     printf("\n");
                 }
                 result.clear();
             }
             path.clear();
             checked.clear();
         }
         for (it = graph.begin(); it != graph.end(); ++it) {
             (it->second).clear();
         }
         graph.clear();
         clearTree(root);
     }

     return ;
 }