LeetCode Binary Tree Paths

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

	}
    
    
};