Given two values k1 and k2 (where k1 < k2) and a root pointer to a Binary Search Tree. Find all the keys of tree in range k1 to k2. i.e. print all x such that k1<=x<=k2 and x is a key of given BST. Return all the keys in ascending order.
Example
If k1 = 10 and k2 = 22, then your function should return [12, 20, 22].
20
/ \
8 22
/ \
4 12
Solution
While 循环：
"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution:
"""
@param root: The root of the binary search tree.
@param k1 and k2: range k1 to k2.
@return: Return all keys that k1<=key<=k2 in ascending order.
"""
def searchRange(self, root, k1, k2):
# write your code here
if root == None:
return []
self.result = []
self.traverse(root, k1, k2)
return self.result
def traverse(self, root, k1, k2):
if root == None:
return
if root.val > k1:
self.traverse(root.left,k1, k2)
if k1 <= root.val and root.val <= k2:
self.result.append(root.val)
if root.val < k2:
self.traverse(root.right,k1, k2)
DFS 循环方法
class Solution:
"""
@param root: The root of the binary search tree.
@param k1 and k2: range k1 to k2.
@return: Return all keys that k1<=key<=k2 in ascending order.
"""
def searchRange(self, root, k1, k2):
# write your code here
if root == None:
return []
stack = []
curr = root
result = []
while stack or curr:
while curr:
stack.append(curr)
curr = curr.left
curr = stack.pop()
if curr.val >= k1 and curr.val <= k2:
result.append(curr.val)
curr = curr.right
return result