# 剑指offer34：二叉树中和为某一值的路径

![在这里插入图片描述](https://img-blog.csdnimg.cn/20190707104409366.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2phY2tpZV9vMm8y,size_16,color_FFFFFF,t_70) 1.判断是不是空树 2.递归结束的条件是叶子节点 3.考察二叉树前序遍历

```python
# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    # 返回二维列表，内部每个列表表示找到的路径
    def FindPath(self, root, expectNumber):
        # write code here
        if not root:
            return []
        if root and not root.left and not root.right and root.val == expectNumber:
            return [[root.val]]
        res = []
        left = self.FindPath(root.left, expectNumber-root.val)
        right = self.FindPath(root.right, expectNumber-root.val)
        for i in left+right:
            res.append([root.val]+i)
        return res
```


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