problem:
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
.
thinking:
(1)思路比较清晰:DFS遍历所有根节点到叶节点的路径,采用string类型保存较方便,最后再将string类型转换为int类型
(2)string类型表示的int数有可能溢出,需要额外注意,这道题的验证程序没有涉及。
code:
class Solution { vector<string> str; public: int sumNumbers(TreeNode *root) { if(root==NULL) return 0; string s; dfs(root,s); int ret=0; for(int i=0;i<str.size();i++) ret+=string_to_int(str[i]); return ret; } void dfs(TreeNode *node, string s) { s.push_back('0'+node->val); if(node->left==NULL && node->right==NULL) { str.push_back(s); return; } if(node->left!=NULL) dfs(node->left,s); if(node->right!=NULL) dfs(node->right,s); } int string_to_int(string s) { int n=s.size(); if(n==0) return 0; int num=0; int a=1; for(int i=n-1;i>=0;i--) { num+=(s.at(i)-'0')*a; a*=10; } return num; } };