LeetCode112.Path Sum

LeetCode113.Path Sum II
LeetCode437.Path Sum III
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.

需要注意的是，在这里路径的起点必须是根节点，终点必须是叶节点。如果中间某节点已经达到总和要求，也必须继续向下遍历。

Code

Recursive

# 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: TreeNode, sum: int) -> bool:
        if root:
            sum -= root.val
            if sum == 0 and self.isLeaf(root):
                return True
            else:
                return self.hasPathSum(root.left, sum) or self.hasPathSum(root.right, sum)

            
    def isLeaf(self, root):
        if root:
            if root.left is None and root.right is None:
                return True
        return False
            

        

一个错误的思想是在达到sum值后直接通过判断是否为叶节点就进行返回，这样的话就是属于我们前面说的终点不一定是叶节点的情况，也有可能是中间节点已经满足sum值，这时候直接返回其是否为叶节点，如果不是则直接被判为False，这就忽略了在其之后的节点和为0的情况。

if sum == 0 :
    return self.isLeaf(root)
else:
    return self.hasPathSum(root.left, sum) or self.hasPathSum(root.right, sum)

在这里要区别局部函数和全局类函数，一个需要在调用时候加self.function一个不需要；

另外需要区别开node和root，sum和target

# 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: TreeNode, sum: int) -> bool:
        if not root:
            return False
        
        def update(node, target):
            if not node:
                return False
            target -= node.val
            if target == 0 and self.isLeaf(node):
                return True
            else:
                return update(node.left, target) or update(node.right, target)
            
        return update(root, sum)
            
            
    def isLeaf(self, root):
        if root:
            if root.left is None and root.right is None:
                return True
        return False

本文地址：https://blog.csdn.net/qq_41704019/article/details/107449738