Leetcode 257.二叉树的所有路径

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257. 二叉树的所有路径

题目

给定一个二叉树，返回所有从根节点到叶子节点的路径。
说明: 叶子节点是指没有子节点的节点。

示例:

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输入:

   1
 /   \
2     3
 \
  5

输出: ["1->2->5", "1->3"]

解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3

方法

方法1：递归

递归先序遍历

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class Solution(object):
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        if not root:
            return []
        if not root.left and not root.right:
            return [str(root.val)]
        left_paths=self.binaryTreePaths(root.left)
        right_paths=self.binaryTreePaths(root.right)
        return ['%s->%s' %(root.val,path) for path in left_paths+right_paths]

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# 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 []

        if not root.left and not root.right:
            return [str(root.val)]

        paths=[str(root.val)+'->'+path for path in self.binaryTreePaths(root.left)]
        paths+=[str(root.val)+'->'+path for path in self.binaryTreePaths(root.right)]
        return paths
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# 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 []

        res=[]
        self.dfs(root,'',res)
        return res

    def dfs(self,root,ls,res):
        if not root.left and not root.right:
            res.append(ls+str(root.val))
        if root.left:
            self.dfs(root.left,ls+str(root.val)+'->',res)
        if root.right:
            self.dfs(root.right,ls+str(root.val)+'->',res)
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# 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]
        """
        self.path=[]
        path=''

        if not root:
            return self.path
        self.road(root,path)
        return self.path
    def road(self,root,path):
        path+=str(root.val)+'->'
        if root.left:
            self.road(root.left,path)
        if root.right:
            self.road(root.right,path)
        if not root.left and not root.right:
            self.path.append(path[:-2])

方法2：迭代

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# 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 []

        res=[]
        stack=[(root,'')]
        while stack:
            node,s=stack.pop()
            if not node.left and not node.right:
                res.append(s+str(node.val))

            if node.right:
                stack.append((node.right,s+str(node.val)+'->'))
            if node.left:
                stack.append((node.left,s+str(node.val)+'->'))
        return res

参考资料