题目如下：
Given two binary search trees, return
Trueif and only if there is a node in the first tree and a node in the second tree whose values sum up to a given integertarget.
Example 1:
Input: root1 = [2,1,4], root2 = [1,0,3], target = 5 Output: true Explanation: 2 and 3 sum up to 5.Example 2:
Input: root1 = [0,-10,10], root2 = [5,1,7,0,2], target = 18 Output: false
Note:
- Each tree has at most
5000nodes.-10^9 <= target, node.val <= 10^9
解题思路：我用的是最直接的方法，把两棵树的节点的值分别存到两个字典中，然后遍历字典，看看能不能组成target。
代码如下：
# Definition for a binary tree node. class TreeNode(object): def __init__(self, x): self.val = x self.left = None self.right = None class Solution(object): def twoSumBSTs(self, root1, root2, target): """ :type root1: TreeNode :type root2: TreeNode :type target: int :rtype: bool """ dic1 = {} dic2 = {} def recursive(node,dic): if node != None: dic[node.val] = 1 if node.left != None: recursive(node.left,dic) if node.right != None: recursive(node.right, dic) recursive(root1,dic1) recursive(root2,dic2) for val in dic1.iterkeys(): if target - val in dic2: return True return False
原文：https://www.cnblogs.com/seyjs/p/11629275.html
