《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 }

 

 posted on 2014-03-19 05:15  zhuli19901106  阅读(270)  评论(0)  编辑  收藏  举报