Python, LintCode, 480. 二叉树的所有路径

最新推荐文章于 2022-12-06 14:31:25 发布

原创最新推荐文章于 2022-12-06 14:31:25 发布 · 364 阅读

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本文介绍了一种用于寻找二叉树所有根节点到叶子节点路径的算法实现。通过递归与迭代两种方式，详细展示了如何获取这些路径并以字符串形式返回。适用于需要解决二叉树相关问题的开发者。

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class Solution:
    """
    @param root: the root of the binary tree
    @return: all root-to-leaf paths
    """
    def binaryTreePaths(self, root):
        # write your code here
        if root == None:
            return []
        nodelist = []
        nodelist.append(root)
        res = []
        res.append(str(root.val))
        res2 = []
        while len(nodelist) != 0:
            root = nodelist.pop(0)
            temp = res.pop(0)
            if root.left == None and root.right == None:
                res2.append(temp)
            if root.left:
                nodelist.append(root.left)
                templ = temp + "->" + str(root.left.val)
                res.append(templ)
            if root.right:
                nodelist.append(root.right)
                tempr = temp + "->" + str(root.right.val)
                res.append(tempr)
        return res2

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class Solution:
    """
    @param root: the root of the binary tree
    @return: all root-to-leaf paths
    """
    def binaryTreePaths(self, root):
        # write your code here
        if root == None:
            return []
        res = []
        if root.left == None and root.right == None:
            return [str(root.val)]
        if root.left:
            templ = self.binaryTreePaths(root.left)
            resl = [str(root.val) + "->" + templ[i] for i in range(len(templ))]
            res += resl
        if root.right:
            tempr = self.binaryTreePaths(root.right)
            resr = [str(root.val) + "->" + tempr[i] for i in range(len(tempr))]
            res += resr
        return res