Problem description:

Given a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

Example:

Input:

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   1
 /   \
2     3
 \
  5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

Solution:

The idea is to use dfs to find every leaf. The definition for leaf is it does not have left child or right child.

DFS

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def binaryTreePaths(self, root: TreeNode) -> List[str]:
        def dfs(root, tmp, res):
            if not root.left and not root.right:
                res.append(tmp + str(root.val))
                return
            if root.left: dfs(root.left, tmp+ str(root.val) + "->", res)
            if root.right: dfs(root.right, tmp+ str(root.val)+ "->", res)
        if not root:
            return []
        res = []
        dfs(root, "", res)
        return res

BFS

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class Solution:
    def binaryTreePaths(self, root: TreeNode) -> List[str]:
        # BFS
        if not root:
            return []
        res = []
        q = collections.deque([(root, "")])
        while q:
            node, tmp = q.popleft()
            if not node.left and not node.right:
                res.append(tmp + str(node.val))
            if node.left:
                q.append([node.left, tmp+str(node.val)+"->"])
            if node.right:
                q.append([node.right, tmp+str(node.val)+"->"])
        return res

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/**
 * 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) return vector<string>();
        string tmp= "";
        helper(res, tmp, root);
        return res;
    }
    
    void helper(vector<string>& res, string tmp, TreeNode* root){
        if(!root->left && !root->right) //reach a leaf
            res.push_back(tmp+to_string(root->val));
        if(root->left) helper(res, tmp+to_string(root->val)+"->", root->left);
        if(root->right) helper(res, tmp+to_string(root->val)+"->", root->right);
    }
    
};

time complexity: $O(n)$, because visit every node once
space complexity: $O(logn)$, height of the tree
reference:
https://goo.gl/RvfssD