[LeetCode&Python] Problem 257. Binary Tree Paths

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

Note: A leaf is a node with no children.

Example:

Input:
 
   1
 /   \
2     3
 \
  5
 
Output: ["1->2->5", "1->3"]
 
Explanation: All root-to-leaf paths are: 1->2->5, 1->3
 
Iteration:

# 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):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        if not root:
            return []
        stack=[]
        stack.append((root,str(root.val)))
        paths=[]
        while stack:
            node,path=stack.pop()
            if not node.left and not node.right:
                paths.append(path)
            else:
                if node.left:
                    stack.append((node.left,path+"->"+str(node.left.val)))
                if node.right:
                    stack.append((node.right,path+"->"+str(node.right.val)))
        return paths

　　

Recursion:

# 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):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        ans=[]
        def isleaf(node):
            if not node.right and not node.left:
                return True
            return False
        def findPath(node,path):
            if node:
                path+=str(node.val)
                if isleaf(node):
                    ans.append(path)
                else:
                    path+="->"
                    findPath(node.left,path)
                    findPath(node.right,path)
        path=""
        findPath(root,path)
        return ans