437.Path Sum III（递归！important！）-优快云博客

本文探讨了在二叉树中寻找所有从根节点到任意节点路径，使得路径上节点值之和等于特定值的问题。文章提供了一种递归算法解决方案，通过遍历树的所有可能路径来计算和匹配给定值。

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You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

Return 3. The paths that sum to 8 are:

5 -> 3
5 -> 2 -> 1
-3 -> 11

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

class Solution:
    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: int
        """
        def solve(root,sum): #from root to the child
            if root is None:
                return 0
            return int(root.val==sum) + solve(root.left,sum-root.val) + solve(root.right,sum-root.val)
        if root is None:
            return 0
        return solve(root,sum) + self.pathSum(root.left,sum) + self.pathSum(root.right,sum) #choose root and not choose root

转载于:https://www.cnblogs.com/bernieloveslife/p/9765106.html