LeetCode437. Path Sum III

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本文介绍了一种在二叉树中查找所有路径和等于给定值的方法，使用递归分治策略实现。路径可以始于任意节点，但必须向下延伸。通过实例展示了如何找到满足条件的路径。

题目描述：

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

      10
     /  \
    5   -3
   / \    \
  3   2   11
 / \   \
3  -2   1

Return 3. The paths that sum to 8 are:

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

题意是在树中找出满足和为sum的路径，要求路径的方向必须是由高到低。

解题思路：

采用分治Divide And Conquer的方式，将原问题分成子问题来解决。原问题：找出以root节点为根节点的树的和为sum的路径；子问题有三个：找出包含root节点的和为sum的路径、找出以root.left为根节点的子树的和为sum的路径、找出以root.right为根节点的子树的和为sum的路径。递归解决即可。

Python代码：

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

class Solution:
    def __init__(self):
        global res
        res = 0
    def pathSum(self, root: 'TreeNode', sum: 'int') -> 'int':
        if not root:
            return 0
        self.traval(root, sum)
        self.pathSum(root.left, sum)
        self.pathSum(root.right, sum)
        return res
    def traval(self, root, sum):
        global res
        if not root:
            return 
        if root.val == sum:
            res += 1
        sum -= root.val
        self.traval(root.left, sum)
        self.traval(root.right, sum)