Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target.
Input:
5
/ \
3 6
/ \ \
2 4 7
Target = 9
Output:
True
Input:
5
/ \
3 6
/ \ \
2 4 7
Target = 28
Output:
False
Inorder traverse the BST, the result will be a sorted array. Then use the same strategy as in two sum - sorted array (using two pointers).
# 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 findTarget(self, root, k):
"""
:type root: TreeNode
:type k: int
:rtype: bool
"""
def traverse(root, ret):
if root == None:
return ret
traverse(root.left, ret)
ret.append(root.val)
traverse(root.right, ret)
ret = []
traverse(root, ret)
left = 0
right = len(ret) - 1
while left < right:
if ret[left] + ret[right] == k:
return True
elif ret[left] + ret[right] < k:
left += 1
else:
right -= 1
return False