《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：

  1 // 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.

  2 #include <cstdio>

  3 #include <unordered_map>

  4 #include <unordered_set>

  5 using namespace std;

  6 

  7 struct TreeNode {

  8     int val;

  9     TreeNode *left;

 10     TreeNode *right;

 11     

 12     TreeNode(int _val = 0):val(_val), left(nullptr), right(nullptr) {};

 13 };

 14 

 15 void constructTree(TreeNode *&root)

 16 {

 17     int val;

 18     

 19     scanf("%d", &val);

 20     if (val == 0) {

 21         root = nullptr;

 22     } else {

 23         root = new TreeNode(val);

 24         constructTree(root->left);

 25         constructTree(root->right);

 26     }

 27 }

 28 

 29 void constructGraph(TreeNode *root, unordered_map<TreeNode *, unordered_set<TreeNode *> > &graph)

 30 {

 31     if (root->left != nullptr) {

 32         graph[root].insert(root->left);

 33         graph[root->left].insert(root);

 34         constructGraph(root->left, graph);

 35     }

 36     

 37     if (root->right != nullptr) {

 38         graph[root].insert(root->right);

 39         graph[root->right].insert(root);

 40         constructGraph(root->right, graph);

 41     }

 42 }

 43 

 44 void DFS(TreeNode *node, vector<TreeNode *> &path, int sum, const int target, 

 45          unordered_set<TreeNode *> &checked, 

 46          unordered_map<TreeNode *, unordered_set<TreeNode *> > &graph, vector<vector<TreeNode *> > &result)

 47 {

 48     path.push_back(node);

 49     checked.insert(node);

 50     

 51     if (sum == target) {

 52         result.push_back(path);

 53     }

 54     unordered_set<TreeNode *>::iterator it;

 55     for (it = graph[node].begin(); it != graph[node].end(); ++it) {

 56         if (checked.find(*it) == checked.end()) {

 57             DFS(*it, path, sum + (*it)->val, target, checked, graph, result);

 58         }

 59     }

 60     

 61     checked.erase(node);

 62     path.pop_back();

 63 }

 64 

 65 void doDFS(TreeNode *root, vector<TreeNode *> &path, const int target, 

 66          unordered_set<TreeNode *> &checked, 

 67          unordered_map<TreeNode *, unordered_set<TreeNode *> > &graph, vector<vector<TreeNode *> > &result)

 68 {

 69     path.clear();

 70     checked.clear();

 71     DFS(root, path, root->val, target, checked, graph, result);

 72     if (root->left != nullptr) {

 73         doDFS(root->left, path, target, checked, graph, result);

 74     }

 75     if (root->right != nullptr) {

 76         doDFS(root->right, path, target, checked, graph, result);

 77     }

 78 }

 79 

 80 void clearTree(TreeNode *&root)

 81 {

 82     if (root == nullptr) {

 83         return;

 84     }

 85     

 86     if (root->left != nullptr) {

 87         clearTree(root->left);

 88     }

 89     if (root->right != nullptr) {

 90         clearTree(root->right);

 91     }

 92     delete root;

 93     root = nullptr;

 94 }

 95 

 96 int main()

 97 {

 98     int i, j;

 99     int target;

100     TreeNode *root;

101     unordered_map<TreeNode *, unordered_set<TreeNode *> > graph;

102     unordered_map<TreeNode *, unordered_set<TreeNode *> >::iterator it;

103     unordered_set<TreeNode *> checked;

104     vector<TreeNode *> path;

105     vector<vector<TreeNode *> > result;

106     

107     while (true) {

108         constructTree(root);

109         if (root == nullptr) {

110             break;

111         }

112         constructGraph(root, graph);

113         while (scanf("%d", &target) == 1 && target) {

114             doDFS(root, path, target, checked, graph, result);

115             if (result.empty()) {

116                 printf("No path is found.\n");

117             } else {

118                 for (i = 0; i < (int)result.size(); ++i) {

119                     printf("%d", result[i][0]->val);

120                     for (j = 1; j < (int)result[i].size(); ++j) {

121                         printf("->%d", result[i][j]->val);

122                     }

123                     result[i].clear();

124                     printf("\n");

125                 }

126                 result.clear();

127             }

128             path.clear();

129             checked.clear();

130         }

131         for (it = graph.begin(); it != graph.end(); ++it) {

132             (it->second).clear();

133         }

134         graph.clear();

135         clearTree(root);

136     }

137     

138     return 0;

139 }