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Given a binary tree, return all root-to-leaf paths.
For example, given the following binary tree:
1 / 2 3 5
All root-to-leaf paths are:
["1->2->5", "1->3"]
采用的是深度遍历，而且是用递归的方法。也可以用栈，不过递归的方法简单，但是耗时长
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<string> binaryTreePaths(TreeNode* root) {
vector<string> svec;
string str;
if (root == NULL)return svec;
dfs(root, str, svec);
return svec;
}
void dfs(TreeNode *root, string str, vector<string> &svec)
{
if(root==nullptr) return;
if (root->left == NULL&&root->right == NULL)
{
str+=to_string(root->val);
svec.push_back(str); //如果遍历到叶子节点，则将这个路径放到容器中去
return;
}
dfs(root->left, str+to_string(root->val)+"->", svec);//遍历左子树
dfs(root->right, str+to_string(root->val)+"->", svec);//遍历右子树
}
};
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原文地址：http://www.cnblogs.com/csudanli/p/5324717.html
