Leetcode 129 Sum Root to Leaf Numbers

  1 #include <iostream>
  2 #include <list>
  3 #include <queue>
  4 using namespace std;
  5 
  6 
  7 struct TreeNode {
  8 int val;
  9 TreeNode *left;
 10 TreeNode *right;
 11 TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 12 };
 13 
 14 struct ListNode {
 15 int val;
 16 ListNode *next;
 17 ListNode(int x) : val(x), next(NULL){}
 18 };
 19 
 20  
 21 
 22 ListNode* create_list(int* array,int n)
 23 {
 24 if (n==0)
 25 return NULL;
 26 ListNode *head,*current;
 27 head = current = new ListNode(array[0]);
 28 for (int i=1;i<n;i++) 
 29 {
 30 ListNode *node = new ListNode(array[i]);
 31 current -> next = node; 
 32 current = current -> next; 
 33 }
 34 return head; 
 35 }
 36 
 37 TreeNode *constructTree(string *dat , int len)
 38 {
 39 if(dat == NULL)
 40 return NULL;
 41 TreeNode *root = NULL;
 42 int index = 0;
 43 if(len > 0)
 44 root = new TreeNode(stoi(dat[index]));
 45 else
 46 return NULL;
 47 list<TreeNode *> node; //其实相当于 queue
 48 node.push_back(root);
 49 index ++;
 50 while(index < len)
 51 {
 52 if(!node.empty())
 53 {
 54 TreeNode *root = node.front();
 55 if(index < len )
 56 {
 57 if(dat[index].compare("#") != 0)
 58 {
 59 root->left = new TreeNode(stoi(dat[index])); 
 60 node.push_back(root->left);
 61 }
 62 index ++;
 63 }
 64 if(index < len )
 65 {
 66 if(dat[index].compare("#") != 0)
 67 {
 68 root->right = new TreeNode(stoi(dat[index])); 
 69 node.push_back(root->right);
 70 }
 71 index ++;
 72 }
 73 node.pop_front();
 74 }
 75 }
 76 return root;
 77 }
 78 
 79 void print(TreeNode * root)
 80 {
 81 if(root == NULL)
 82 return ;
 83 queue <TreeNode *> q;
 84 q.push(root);
 85 while(!q.empty())
 86 {
 87 TreeNode *cur = q.front();
 88 cout<< cur->val; 
 89 q.pop();
 90 if(cur -> left) q.push(cur->left);
 91 if(cur -> right) q.push(cur->right);
 92 }
 93 }
 94 
 95 
 96 class Solution {
 97 public:
 98 int sumNumbers(TreeNode* root) { // DFS 想好递归的参数和返回值
 99 if (root == NULL)
100 return 0;
101 if (root -> left == NULL && root -> right == NULL)
102 return root -> val; 
103 int total = root -> val;
104 return dfs(root, 0) ;
105 }
106 int dfs(TreeNode* root, int sum)
107 {
108 if(root->left == NULL && root->right == NULL)
109 {
110 return sum*10+root->val;
111 }
112 
113 else if (root ->left != NULL && root -> right != NULL)
114 return dfs(root->left, root->val + sum*10) + dfs(root->right, root->val + sum*10);
115 else if (root -> right != NULL)
116 return dfs(root->right, root->val + sum*10);
117 else if (root -> left != NULL)
118 return dfs(root->left, root->val + sum*10);
119 }
120 };
121 
122 int main()
123 {
124 //vector <int> a = {1,3,5};
125 Solution ans;
126 //string a[13] = {"5","4","8","11","#","13","4","7","1","#","#","#","1"};
127 string a[5] = {"1","0","#","#","#"};
128 
129 TreeNode * tree = constructTree(a,5);
130 int result = ans.sumNumbers(tree);
131 cout<<result;
132 
133 //for ()
134 //ListNode * pure = ans.addTwoNumbers(first,second);
135 //if(pure == NULL)
136 // printf("NULL");
137 //else
138 //print(pure);
139 return 0;
140 }