0257-二叉树的所有路径

给你一个二叉树的根节点 root ，按 任意顺序 ，返回所有从根节点到叶子节点的路径。

叶子节点 是指没有子节点的节点。

示例 1：

输入：root = [1,2,3,null,5]
输出：["1->2->5","1->3"]
示例 2：

输入：root = [1]
输出：["1"]

提示：

树中节点的数目在范围 [1, 100] 内
-100 <= Node.val <= 100

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/binary-tree-paths

参考：

• https://leetcode-cn.com/problems/binary-tree-paths/solution/er-cha-shu-de-suo-you-lu-jing-by-leetcode-solution/
• https://leetcode-cn.com/problems/binary-tree-paths/solution/dai-ma-sui-xiang-lu-dai-ni-xue-tou-er-ch-cxmu/

python

# 0257.二叉树的所有路径
class Solution:
    def binaryTreePaths(self, root: TreeNode) -> [str]:
        path = []
        res = []
        def backtrace(root, path):
            if not root:
                return
            path.append(root.val)
            if not root.left and not root.right:
                res.append(path[:])
            ways = []
            if root.left:
                ways.append(root.left)
            if root.right:
                ways.append(root.right)
            for way in ways:
                backtrace(way, path)
                path.pop()
        backtrace(root, path)
        return [ "->".join(list(map(str, i))) for i in res ]

    def binaryTreePaths_DFS(self, root: TreeNode) -> [str]:
        def construct_paths(root, path):
            if root:
                path += str(root.val)
                if not root.left and not root.right: # 叶子结点
                    paths.append(path) # 路径加入结果集
                else: # 非叶子节点，继续递归
                    path += "->"
                    construct_paths(root.left, path)
                    construct_paths(root.right, path)
        paths = []
        construct_paths(root, "")
        return paths

    def binaryTreePaths_BFS(self, root: TreeNode) -> [str]:
        paths = []
        if not root:
            return paths
        from collections import deque
        node_queue = deque([root])
        path_queue = deque([str(root.val)])

        while node_queue:
            node = node_queue.popleft()
            path = path_queue.popleft()

            if not node.left and not node.right:
                paths.append(path)
            else:
                if node.left:
                    node_queue.append(node.left)
                    path_queue.append(path + "->" + str(node.left.val))

                if node.right:
                    node_queue.append(node.right)
                    path_queue.append(path + "->" + str(node.right.val))
        return paths

golang

package binaryTree

import "strconv"

func binaryTreePaths(root *TreeNode) []string {
	res := make([]string, 0)
	var travel func(node *TreeNode, s string)
	travel= func(node *TreeNode, s string) {
		if node.Left == nil && node.Right == nil {
			v := s + strconv.Itoa(node.Val)
			res = append(res, v)
			return
		}
		s = s + strconv.Itoa(node.Val) + "->"
		if node.Left != nil {
			travel(node.Left, s)
		}
		if node.Right != nil {
			travel(node.Right, s)
		}
	}
	travel(root, "")
	return res
}

func binaryTreePathsDFS(root *TreeNode) []string {
	paths := []string{}
	construcPaths(root, "")
	return paths
}

func construcPaths(root *TreeNode, path string)  {
	if root != nil {
		pathSB := path
		pathSB += strconv.Itoa(root.Val)
		if root.Left == nil && root.Right == nil {
			paths = append(paths, pathSB)
		} else {
			pathSB += "->"
			construcPaths(root.Left, pathSB)
			construcPaths(root.Right, pathSB)
		}
	}
}

func binaryTreePathsBFS(root *TreeNode) []string {
	paths := []string{}
	if root == nil {
		return paths
	}
	nodeQueue := []*TreeNode{}
	pathQueue := []string{}
	nodeQueue = append(nodeQueue, root)
	pathQueue = append(pathQueue, strconv.Itoa(root.Val))

	for i:=0;i" +strconv.Itoa(node.Left.Val))
		}
		if node.Right != nil {
			nodeQueue =append(nodeQueue, node.Right)
			pathQueue = append(pathQueue, path + "->" +strconv.Itoa(node.Right.Val))
		}
	}
	return paths
}