leetcode112 -- Path Sum

leetcode112 -- Path Sum

题目


Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

就是从root到根要找到一条路径使得沿着这条路径的和等于一个给定的数。 这道题我想的是用递归来求解,每往下一次就更新这个和.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:      
        return self.helper(root, sum)
    
    def helper(self, root, target):
        if not root:
            return False
        if not root.left and not root.right:
            if root.val == target:
                return True
            else:
                return False
        
        if root.left and not root.right:
            return self.helper(root.left, target-root.val)
        if not root.left and root.right:
            return self.helper(root.right, target-root.val)
        
        if root.left and root.right:
            return self.helper(root.left, target-root.val) or self.helper(root.right, target-root.val)
            

可以改成下面好看一些的

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:      
        return self.helper(root, sum)
    
    def helper(self, root, target):
        if not root:
            return False
        
        if not root.left and not root.right:
            if root.val == target:
                return True
            else:
                return False
        
        elif root.left and not root.right:
            return self.helper(root.left, target-root.val)
        elif not root.left and root.right:
            return self.helper(root.right, target-root.val)
        else: 
            return self.helper(root.left, target-root.val) or self.helper(root.right, target-root.val)
            

也可以用下面的办法,即先得到所有的从根节点到叶子节点的路径,然后再判断。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:      
        res = []
        self.helper(root, res, [])
        
       
        # get res
        print(res)
        for x in res:
            summ =0
            for y in x:
                summ += y
            if summ == sum:
                return True
            
        return False
    
    
    def helper(self, root, res, out):
        if not root:
            return 
        
        out.append(root.val)
        
        if not root.left and not root.right:
            res.append(out[:])
            
        self.helper(root.left, res, out[:])
        self.helper(root.right, res, out[:])
        

打赏,谢谢~~

取消

感谢您的支持,我会继续努力的!

扫码支持
扫码打赏,多谢支持~

打开微信扫一扫,即可进行扫码打赏哦