给你二叉树的根节点 root 和一个表示目标和的整数 targetSum 。 判断该树中是否存在 根节点到叶子节点 的路径,这条路径上所有节点值相加等于目标和 targetSum 。 如果存在,返回 true ;否则,返回 false 。 叶子节点 是指没有子节点的节点示例 1: 输入:root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22 输出:true 解释:等于目标和的根节点到叶节点路径如上图所示。 示例 2: 输入:root = [1,2,3], targetSum = 5 输出:false 解释:树中存在两条根节点到叶子节点的路径: (1 --> 2): 和为 3 (1 --> 3): 和为 4 不存在 sum = 5 的根节点到叶子节点的路径。 示例 3: 输入:root = [], targetSum = 0 输出:false 解释:由于树是空的,所以不存在根节点到叶子节点的路径。提示: 树中节点的数目在范围 [0, 5000] 内 -1000 <= Node.val <= 1000 -1000 <= targetSum <= 1000
解题思路:【DFS + 回溯】
1、Python输入[5,4,8,11,null,13,4,7,2,null,null,null,1]无法直接表达
可替换为[5,4,8,11,None,13,4,7,2,None,None,None,1]
2、根据层序遍历构建二叉树
3、DFS
DFS:前序遍历;中-左-右将路径节点、targetSum作为参数传入DFS函数处理中节点:当路径节点总和 == targetSum 且当前节点为叶子节点时,找到路径处理左节点:DFS遍历左节点处理右节点:DFS遍历右节点
class TreeNode:def __init__(self, val, left=None, right=None):self.val = valself.left = leftself.right = right
class Solution:def build_tree(self, levelorder=[]):length = len(levelorder)if length == 0:return Noneroot = TreeNode(levelorder[0])queue = [root]i = 1while queue and i < length:node = queue.pop(0)if levelorder[i] != None:node.left = TreeNode(levelorder[i])queue.append(node.left)i += 1if i < length and levelorder[i] != None:node.right = TreeNode(levelorder[i])queue.append(node.right)i += 1return rootdef dfs(self,root, targetSum, res, tmpSum):# 树为空,直接返回Falseif not root:res[0] = False# 找到叶子节点,且路径和 = targetSumif sum(tmpSum) == targetSum and not root.left and not root.right:res[0] = Trueif root.left:self.dfs(root.left, targetSum, res, tmpSum + [root.left.val])if root.right:self.dfs(root.right, targetSum, res, tmpSum + [root.right.val])if __name__ == '__main__':try:nums = eval(input())solution =Solution()root = solution.build_tree(nums)targetSum = int(input())res = [False]if root == None:print(False)else:solution.dfs(root, targetSum, res, [root.val])print(res[0])except Exception as e:print(e)
其他思路:【BFS】
BFS:使用队列保存每个节点路径和若存在一个节点恰好是叶子节点,且路径和 = targetSum,找到路径
def bfs(self, root, res, targetSum):if not root:return res[0]queue = [(root, root.val)]while queue:node, path = queue.pop(0)if not node.left and not node.right and path == targetSum:res[0] = Truereturn res[0] # 找到即返回,减少遍历次数if node.left:queue.append((node.left, path + node.left.val))if node.right:queue.append((node.right, path + node.right.val))return res[0]
仅作为代码记录,方便自学自查自纠