Leetcode 257. Binary Tree Paths

题目描述：返回二叉树从根到叶子节点的路径。（是无左子树且无右子树）

题目链接：Leetcode 257. Binary Tree Paths

递归添加。

代码如下

class Solution:
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        ans = []
        if not root:
            return []
        self.dfs(root,[],ans)
        return ans
    
    def dfs(self,root,path,ans):
        if root and not root.left and not root.right:
            path.append(str(root.val))
            tmp = "->".join(path)
            if tmp not in ans:
                ans.append(tmp)
            return 
        if root:
            self.dfs(root.left,path+[str(root.val)],ans)
            self.dfs(root.right,path+[str(root.val)],ans)

非递归

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

class Solution:
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        if not root:
            return []
        ans = []
        s = [(root,[str(root.val)])]
        while s:
            node,path = s.pop()
            
            if not node.left and not node.right:
                # path.append(str(node.val))
                ans.append("->".join(path))
            if node.right:
                # path.append()
                p = list(path)
                p.append(str(node.right.val))
                s.append((node.right,p))
            if node.left:
                p = list(path)
                p.append(str(node.left.val))
                s.append((node.left,p))
                # s.append((node.left,path+[str(node.left.val)]))
        return ans

class Solution:
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        if not root:
            return []
        
        paths = []
        stack = [(root, str(root.val))]
        while stack:
            node, path = stack.pop()
            if not node.left and not node.right:
                paths.append(path)
            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

参考链接

• Leetcode 257. Binary Tree Paths直接用字符串，每次+都会返回新的字符串。