Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path 1->2->3 which represents the number 123.
Find the total sum of all root-to-leaf numbers.
For example,
1 / 2 3
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Return the sum = 12 + 13 = 25
思路
递归。每一层如果有子节点的话，就把之前路径的值往下传递，直到到达叶节点加到最终结果里。
1 int result,tmp; 2 void calSum(TreeNode *root){ 3 if(root == NULL) 4 return; 5 if(root->left == NULL && root->right == NULL){ 6 tmp = tmp*10+(root->val); 7 result += tmp; 8 tmp /= 10; 9 return; 10 } 11 if(root->left != NULL){ 12 tmp = tmp*10+(root->val); 13 calSum(root->left); 14 tmp /= 10; 15 } 16 if(root->right != NULL){ 17 tmp = tmp*10+(root->val); 18 calSum(root->right); 19 tmp /= 10; 20 } 21 } 22 int sumNumbers(TreeNode *root) { 23 // Note: The Solution object is instantiated only once and is reused by each test case. 24 result = 0; 25 tmp = 0; 26 calSum(root); 27 return result; 28 }
第二次代码
1 void search(TreeNode* root, int &result, int last){ 2 if(root){ 3 if(!root->left && !root->right) 4 result += (last*10+root->val); 5 else{ 6 last = last*10+root->val; 7 search(root->left, result, last); 8 search(root->right, result, last); 9 } 10 } 11 } 12 int sumNumbers(TreeNode *root) { 13 int result = 0; 14 search(root, result, 0); 15 return result; 16 }