Search Range in Binary Search Tree

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