257. Binary Tree Paths（列印二叉樹所有路徑）

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"]

首先我採用了dfs，但是我的方法有點麻煩。

方法一：dfs

/**
 * 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> res;
        if(root == NULL)
            return res;
        string str = std::to_string(root->val);
        if(root->left != NULL || root->right != NULL){
            dfs(root->left, res, str);
            dfs(root->right, res, str);
        }
        else 
            res.push_back(str);
        return res;
    }
    void dfs(TreeNode* root, vector<string>& res, string str){
        if(root == NULL) return;
        str += "->" + std::to_string(root->val);
        if(root->left == NULL && root->right == NULL){
            res.push_back(str);
            return ;
        }
        dfs(root->left, res, str);
        dfs(root->right, res, str);
    }
};

然後我參照別人的dfs，重新用Python寫了一遍，簡化了一些步驟。

下面是dfs的簡潔版本：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def binaryTreePaths(self, root):
        if not root:
            return []
        res = []
        self.dfs(root, res, "")
        return res;

    def dfs(self, root, res, ls):
        if not root.left and not root.right:
            res.append(ls+str(root.val));
            return res
        if root.left:
            self.dfs(root.left, res, ls+str(root.val)+"->")
        if root.right:
            self.dfs(root.right, res, ls+str(root.val)+"->")

非遞迴版的dfs，類似於二叉樹的前序遍歷，使用stack：

class Solution(object):
    def binaryTreePaths(self, root):
        res = []
        if not root:
            return res
        self.dfs(root, res)
        return res;

    def dfs(self, root, res):
        if not root:
            return []
        stack = [(root, "")]
        while stack:
            node, ls = stack.pop()
            if not node.left and not node.right:
                res.append(ls+str(node.val))
            if node.right:
                stack.append((node.right, ls+str(node.val)+"->"))
            if node.left:
                stack.append((node.left, ls+str(node.val)+"->"))
        return res

方法二：採用廣度優先BFS，使用佇列：

class Solution(object):
    def binaryTreePaths(self, root):
        res = []
        if not root:
            return res
        self.bfs(root, res)
        return res;

    def bfs(self, root, res):
        if not root:
            return []
        queue = collections.deque([(root, "")])
        while queue:
            node, ls = queue.popleft()
            if not node.left and not node.right:
                res.append(ls+str(node.val))
            if node.left:
                queue.append((node.left, ls+str(node.val)+"->"))
            if node.right:
                queue.append((node.right, ls+str(node.val)+"->"))
        return res