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  1. Path Sum

简单

https://leetcode-cn.com/problems/path-sum/

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 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

相关企业

  • Facebook|6
  • 字节跳动|5
  • 亚马逊 Amazon|5
  • 甲骨文 Oracle|3

相关标签

  • Tree
  • Depth-First Search
  • Breadth-First Search
  • Binary Tree

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# 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

        resultCtr = []
        self._dfs(root, 0, targetSum, resultCtr)
        return len(resultCtr) > 0

    def _dfs(self, currnode, currpathSum, targetSum, resultCtr):
        if not currnode:
            return 

        currpathSum += currnode.val
        # print(currnode.val, currpathSum, resultCtr, targetSum)

        # leaf
        if not currnode.left and not currnode.right and currpathSum == targetSum:
            resultCtr.append(1)
        # print(currnode.val, currpathSum, resultCtr, targetSum)

        # non-leaf
        self._dfs(currnode.left, currpathSum, targetSum, resultCtr)
        self._dfs(currnode.right, currpathSum, targetSum, resultCtr)
        currpathSum -= currnode.val
# 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
        elif not root.left and not root.right and root.val == targetSum:
            return True
        else:
            return self.hasPathSum(root.left, targetSum - root.val) or self.hasPathSum(root.right, targetSum - root.val)