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112. Path Sum ​

题目 ​

Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

A leaf is a node with no children.

 

Example 1:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output: true
Explanation: The root-to-leaf path with the target sum is shown.

Example 2:

Input: root = [1,2,3], targetSum = 5
Output: false
Explanation: There are two root-to-leaf paths in the tree:
(1 --> 2): The sum is 3.
(1 --> 3): The sum is 4.
There is no root-to-leaf path with sum = 5.

Example 3:

Input: root = [], targetSum = 0
Output: false
Explanation: Since the tree is empty, there are no root-to-leaf paths.

 

Constraints:

  • The number of nodes in the tree is in the range [0, 5000].
  • -1000 <= Node.val <= 1000
  • -1000 <= targetSum <= 1000
Related Topics
  • 树
  • 深度优先搜索
  • 广度优先搜索
  • 二叉树

  • 👍 1459
  • 👎 0

    • 思路 ​

    注意本题是寻找根到叶子节点的距离

    叶子节点的终止条件是 not root.left and not root.right

    而不是根节点的终止条件 not root

    解法 ​

    py
    # leetcode submit region begin(Prohibit modification and deletion)
# 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 hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
        if not root:
            return False
        if not root.left and not root.right:
            if targetSum == root.val:
                return True

        return self.hasPathSum(root.left, targetSum - root.val) or self.hasPathSum(root.right, targetSum - root.val)


        
# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    N = sum of treeNode H = height of tree

    • 时间复杂度 O(N)
    • 空间复杂度 O(H)