给定一个二叉树，返回所有从根节点到叶子节点的路径。
说明: 叶子节点是指没有子节点的节点。
struct Node { int value; Node* left; Node* right; }; vector<string> printRoutes(Node* root) { stack<Node*> s; s.push(root); string str = ""; vector<string> ret; while(!s.empty()) { root = s.top(); s.pop(); if(root->right) s.push(root->right); if(root->left) s.push(root->left); str += root->value; if(root->left == NULL && root->right == NULL) { ret.push_back(str); str.erase(str.size() - 1 , 1); } } return ret; } int main() { Node a , b , c; a.value = 'a'; a.left = &b; a.right = &c; b.value = 'b'; b.left = NULL; b.right = NULL; c.value = 'c'; c.left = NULL; c.right = NULL; vector<string> ret = printRoutes(&a); for(auto& str:ret) { cout<<str<<endl; } }