113. 路径总和 II

Difficulty
Medium
Tags
二叉树
URL
https://leetcode-cn.com/problems/path-sum-ii/
Star
给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。
叶子节点 是指没有子节点的节点。
示例 1：
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输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：[[5,4,11,2],[5,8,4,5]]
示例 2：
输入：root = [1,2,3], targetSum = 5
输出：[]
示例 3：
输入：root = [1,2], targetSum = 0
输出：[]
提示：
  • 树中节点总数在范围 [0, 5000] 内
  • 1000 <= Node.val <= 1000
  • 1000 <= targetSum <= 1000

分析

本题为
🛵
257. 二叉树的所有路径 和
🕐
112. 路径总和 结合

法1 迭代法

思路
前序遍历二叉树，然后遍历每个节点的时候，将从根节点到该节点的路径值放到列表中，并对应的存放到栈中。然后当访问叶子节点时，判断是否满需条件
题解
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]:
        result = []
        if root is None:
            return result
        stack = []
        stack.append(root)
        stack.append([root.val])
        while len(stack) > 0:
            path_val = stack.pop()
            node = stack.pop()
            if node.left is None and node.right is None:
                if sum(path_val) == targetSum:
                    result.append(path_val)
            if node.left:
                stack.append(node.left)
                stack.append(path_val+[node.left.val])
            if node.right:
                stack.append(node.right)
                stack.append(path_val+[node.right.val])
        return result

递归法（回溯法）

思路
整体思路桶迭代法，只是将存到栈中的路径当做参数传给下一层了
题解
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]:
        self.targetSum = targetSum
        self.result = []
        if root is None:
            return self.result
        self.dfs(node=root, path=[root.val])
        return self.result

    def dfs(self, node, path):
        if node.left is None and node.right is None:
            if sum(path) == self.targetSum:
                self.result.append(path)
        
        if node.left:
            self.dfs(node.left, path+[node.left.val])
        if node.right:
            self.dfs(node.right, path+[node.right.val])
        
        return