BST¶
Table of Contents¶
- 700. Search in a Binary Search Tree (Easy)
- 98. Validate Binary Search Tree (Medium)
- 530. Minimum Absolute Difference in BST (Easy)
- 501. Find Mode in Binary Search Tree (Easy)
- 235. Lowest Common Ancestor of a Binary Search Tree (Medium)
- 701. Insert into a Binary Search Tree (Medium)
- 450. Delete Node in a BST (Medium)
- 669. Trim a Binary Search Tree (Medium)
- 108. Convert Sorted Array to Binary Search Tree (Easy)
- 109. Convert Sorted List to Binary Search Tree (Medium)
- 538. Convert BST to Greater Tree (Medium)
- 230. Kth Smallest Element in a BST (Medium)
- 173. Binary Search Tree Iterator (Medium)
- 1586. Binary Search Tree Iterator II (Medium) 👑
700. Search in a Binary Search Tree¶
-
LeetCode | LeetCode CH (Easy)
-
Tags: tree, binary search tree, binary tree
Binary Search Tree¶
- Binary Tree
- Left subtree of a node contains only nodes with keys less than the node's key
- Right subtree of a node contains only nodes with keys greater than the node's key
- The left and right subtree each must also be a binary search tree
- There must be no duplicate nodes
- Inorder traversal of a BST gives a sorted list of keys
graph TD
4((4)) --- 2((2))
4 --- 7((7))
2 --- 1((1))
2 --- 3((3))700. Search in a Binary Search Tree - Python Solution
from typing import Optional
from binarytree import build
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
# 1. Recursive
def searchBSTRecursive(
root: Optional[TreeNode], val: int
) -> Optional[TreeNode]:
if not root:
return None
if root.val > val:
return searchBSTRecursive(root.left, val)
elif root.val < val:
return searchBSTRecursive(root.right, val)
else:
return root
# 2. Iterative
def searchBSTIterative(
root: Optional[TreeNode], val: int
) -> Optional[TreeNode]:
while root:
if root.val > val:
root = root.left
elif root.val < val:
root = root.right
else:
return root
return None
root = [4, 2, 7, 1, 3]
val = 2
root = build(root)
print(root)
# __4
# / \
# 2 7
# / \
# 1 3
print(searchBSTRecursive(root, val))
# 2
# / \
# 1 3
print(searchBSTIterative(root, val))
# 2
# / \
# 1 3
98. Validate Binary Search Tree¶
-
LeetCode | LeetCode CH (Medium)
-
Tags: tree, depth first search, binary search tree, binary tree
98. Validate Binary Search Tree - Python Solution
from typing import Optional
from binarytree import Node as TreeNode
from binarytree import build
def isValidBST(root: Optional[TreeNode]) -> bool:
inorder = [] # inorder traversal of BST
def dfs(node):
if not node:
return None
dfs(node.left)
inorder.append(node.val)
dfs(node.right)
dfs(root)
for i in range(1, len(inorder)):
if inorder[i] <= inorder[i - 1]:
return False
return True
root = [5, 1, 4, None, None, 3, 6]
root = build(root)
print(root)
# 5__
# / \
# 1 4
# / \
# 3 6
print(isValidBST(root)) # False
# [1, 5, 3, 4, 6]
98. Validate Binary Search Tree - C++ Solution
#include <cassert>
#include <vector>
using namespace std;
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right)
: val(x), left(left), right(right) {}
};
class Solution {
private:
vector<int> inorder;
bool check(vector<int> inorder) {
int n = inorder.size();
if (n <= 1) return true;
for (int i = 1; i < n; i++) {
if (inorder[i] <= inorder[i - 1]) return false;
}
return true;
}
public:
bool isValidBST(TreeNode *root) {
auto dfs = [&](auto &&self, TreeNode *node) -> void {
if (!node) return;
self(self, node->left);
inorder.push_back(node->val);
self(self, node->right);
};
dfs(dfs, root);
return check(inorder);
}
};
int main() {
Solution s;
TreeNode *root = new TreeNode(2);
root->left = new TreeNode(1);
root->right = new TreeNode(3);
assert(s.isValidBST(root) == true);
root = new TreeNode(5);
root->left = new TreeNode(1);
root->right = new TreeNode(4);
root->right->left = new TreeNode(3);
root->right->right = new TreeNode(6);
assert(s.isValidBST(root) == false);
root = new TreeNode(5);
root->left = new TreeNode(4);
root->right = new TreeNode(6);
root->right->left = new TreeNode(3);
root->right->right = new TreeNode(7);
assert(s.isValidBST(root) == false);
return 0;
}
530. Minimum Absolute Difference in BST¶
-
LeetCode | LeetCode CH (Easy)
-
Tags: tree, depth first search, breadth first search, binary search tree, binary tree
530. Minimum Absolute Difference in BST - Python Solution
from typing import Optional
from binarytree import build
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def getMinimumDifference(root: Optional[TreeNode]) -> int:
inorder = []
result = float("inf")
def dfs(node):
if not node:
return None
dfs(node.left)
inorder.append(node.val)
dfs(node.right)
dfs(root)
for i in range(1, len(inorder)):
result = min(result, abs(inorder[i] - inorder[i - 1]))
return result
root = [4, 2, 6, 1, 3]
root = build(root)
print(root)
# __4
# / \
# 2 6
# / \
# 1 3
print(getMinimumDifference(root)) # 1
501. Find Mode in Binary Search Tree¶
-
LeetCode | LeetCode CH (Easy)
-
Tags: tree, depth first search, binary search tree, binary tree
501. Find Mode in Binary Search Tree - Python Solution
from typing import List, Optional
from binarytree import build
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def findMode(root: Optional[TreeNode]) -> List[int]:
hashmap = dict()
def dfs(node):
if not node:
return None
dfs(node.left)
if node.val not in hashmap:
hashmap[node.val] = 1
else:
hashmap[node.val] += 1
dfs(node.right)
dfs(root)
max_counts = max(hashmap.values())
result = []
for key, value in hashmap.items():
if value == max_counts:
result.append(key)
return result
root = [1, None, 2, None, None, 2]
root = build(root)
print(root)
# 1__
# \
# 2
# /
# 2
print(findMode(root)) # [2]
235. Lowest Common Ancestor of a Binary Search Tree¶
-
LeetCode | LeetCode CH (Medium)
-
Tags: tree, depth first search, binary search tree, binary tree
235. Lowest Common Ancestor of a Binary Search Tree - Python Solution
from binarytree import build
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def lowestCommonAncestor(
root: "TreeNode", p: "TreeNode", q: "TreeNode"
) -> "TreeNode":
while root:
if root.val > p.val and root.val > q.val:
root = root.left
elif root.val < p.val and root.val < q.val:
root = root.right
else:
return root
root = [6, 2, 8, 0, 4, 7, 9, None, None, 3, 5]
root = build(root)
p = root.left
q = root.right
print(root)
# ______6__
# / \
# 2__ 8
# / \ / \
# 0 4 7 9
# / \
# 3 5
print(lowestCommonAncestor(root, p, q))
# ______6__
# / \
# 2__ 8
# / \ / \
# 0 4 7 9
# / \
# 3 5
701. Insert into a Binary Search Tree¶
-
LeetCode | LeetCode CH (Medium)
-
Tags: tree, binary search tree, binary tree
701. Insert into a Binary Search Tree - Python Solution
from typing import Optional
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def insertIntoBST(root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if root is None:
return TreeNode(val)
if root.val > val:
root.left = insertIntoBST(root.left, val)
if root.val < val:
root.right = insertIntoBST(root.right, val)
return root
root = TreeNode(4)
root.left = TreeNode(2)
root.right = TreeNode(6)
root.left.left = TreeNode(1)
root.left.right = TreeNode(3)
# __4
# / \
# 2 6
# / \
# 1 3
insertIntoBST(root, 5)
# __4
# / \
# 2 6
# / \ /
# 1 3 5
450. Delete Node in a BST¶
-
LeetCode | LeetCode CH (Medium)
-
Tags: tree, binary search tree, binary tree
450. Delete Node in a BST - Python Solution
from typing import Optional
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def deleteNode(root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
if root is None:
return root
if root.val == key:
if root.left is None and root.right is None:
return None
elif root.left is None:
return root.right
elif root.right is None:
return root.left
else:
cur = root.right
while cur.left is not None:
cur = cur.left
cur.left = root.left
return root.right
if root.val > key:
root.left = deleteNode(root.left, key)
if root.val < key:
root.right = deleteNode(root.right, key)
return root
root = TreeNode(5)
root.left = TreeNode(3)
root.right = TreeNode(6)
root.left.left = TreeNode(2)
root.left.right = TreeNode(4)
root.right.right = TreeNode(7)
# __5
# / \
# 3 6
# / \ \
# 2 4 7
deleteNode(root, 3)
# __5
# / \
# 4 6
# / \
# 2 7
669. Trim a Binary Search Tree¶
-
LeetCode | LeetCode CH (Medium)
-
Tags: tree, depth first search, binary search tree, binary tree
669. Trim a Binary Search Tree - Python Solution
from typing import Optional
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def trimBST(
root: Optional[TreeNode], low: int, high: int
) -> Optional[TreeNode]:
if root is None:
return None
if root.val < low:
return trimBST(root.right, low, high)
if root.val > high:
return trimBST(root.left, low, high)
root.left = trimBST(root.left, low, high)
root.right = trimBST(root.right, low, high)
return root
root = TreeNode(3)
root.left = TreeNode(0)
root.right = TreeNode(4)
root.left.right = TreeNode(2)
root.left.right.left = TreeNode(1)
# __3
# / \
# 0 4
# \
# 2
# /
# 1
trimBST(root, 1, 3)
# __3
# /
# 2
# /
# 1
108. Convert Sorted Array to Binary Search Tree¶
-
LeetCode | LeetCode CH (Easy)
-
Tags: array, divide and conquer, tree, binary search tree, binary tree
108. Convert Sorted Array to Binary Search Tree - Python Solution
from typing import List, Optional
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def sortedArrayToBST(nums: List[int]) -> Optional[TreeNode]:
if len(nums) == 0:
return None
mid = len(nums) // 2
root = TreeNode(nums[mid])
root.left = sortedArrayToBST(nums[:mid])
root.right = sortedArrayToBST(nums[mid + 1 :])
return root
nums = [-10, -3, 0, 5, 9]
root = sortedArrayToBST(nums)
# 0
# / \
# -3 9
# / /
# -10 5
108. Convert Sorted Array to Binary Search Tree - C++ Solution
#include <vector>
using namespace std;
struct TreeNode {
int val;
TreeNode* left;
TreeNode* right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode* left, TreeNode* right)
: val(x), left(left), right(right) {}
};
class Solution {
public:
TreeNode* sortedArrayToBST(vector<int>& nums) {
if (nums.size() == 0) return nullptr;
int mid = nums.size() / 2;
TreeNode* root = new TreeNode(nums[mid]);
vector<int> left(nums.begin(), nums.begin() + mid);
vector<int> right(nums.begin() + mid + 1, nums.end());
root->left = sortedArrayToBST(left);
root->right = sortedArrayToBST(right);
return root;
}
};
int main() { return 0; }
109. Convert Sorted List to Binary Search Tree¶
-
LeetCode | LeetCode CH (Medium)
-
Tags: linked list, divide and conquer, tree, binary search tree, binary tree
109. Convert Sorted List to Binary Search Tree - Python Solution
from typing import Optional
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def sortedListToBST(head: Optional[ListNode]) -> Optional[TreeNode]:
if not head:
return None
def find_mid(head: ListNode) -> ListNode:
prev = None
slow = head
fast = head
while fast and fast.next:
prev = slow
slow = slow.next
fast = fast.next.next
if prev:
prev.next = None
return slow
mid = find_mid(head)
node = TreeNode(mid.val)
if head == mid:
return node
node.left = sortedListToBST(head)
node.right = sortedListToBST(mid.next)
return node
head = ListNode(-10)
head.next = ListNode(-3)
head.next.next = ListNode(0)
head.next.next.next = ListNode(5)
head.next.next.next.next = ListNode(9)
root = sortedListToBST(head)
assert root.val == 0
assert root.left.val == -3
assert root.left.left.val == -10
assert root.right.val == 9
assert root.right.left.val == 5
print("All passed")
# 0
# / \
# -3 9
# / /
# -10 5
538. Convert BST to Greater Tree¶
-
LeetCode | LeetCode CH (Medium)
-
Tags: tree, depth first search, binary search tree, binary tree
538. Convert BST to Greater Tree - Python Solution
from typing import Optional
from binarytree import build
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def convertBST(root: Optional[TreeNode]) -> Optional[TreeNode]:
prev = 0
def dfs(node):
if not node:
return None
nonlocal prev
dfs(node.right)
node.val += prev
prev = node.val
dfs(node.left)
dfs(root)
return root
root = [4, 1, 6, 0, 2, 5, 7, None, None, None, 3, None, None, None, 8]
root = build(root)
print(root)
# ____4__
# / \
# 1 6
# / \ / \
# 0 2 5 7
# \ \
# 3 8
greater_tree = convertBST(root)
print(greater_tree)
# _______30___
# / \
# _36 _21
# / \ / \
# 36 35 26 15
# \ \
# 33 8
230. Kth Smallest Element in a BST¶
-
LeetCode | LeetCode CH (Medium)
-
Tags: tree, depth first search, binary search tree, binary tree
230. Kth Smallest Element in a BST - Python Solution
from typing import Optional
from binarytree import Node as TreeNode
from binarytree import build
# Recursive
def kthSmallestRecursive(root: Optional[TreeNode], k: int) -> int:
inorder = []
def dfs(node):
if not node:
return None
dfs(node.left)
inorder.append(node.val)
dfs(node.right)
dfs(root)
return inorder[k - 1]
# Iteratve
def kthSmallestIteratve(root: Optional[TreeNode], k: int) -> int:
stack = []
while True:
while root:
stack.append(root)
root = root.left
root = stack.pop()
k -= 1
if not k:
return root.val
root = root.right
root = build([3, 1, 4, None, 2])
k = 1
print(root)
# __3
# / \
# 1 4
# \
# 2
print(kthSmallestRecursive(root, k)) # 1
print(kthSmallestIteratve(root, k)) # 1
173. Binary Search Tree Iterator¶
-
LeetCode | LeetCode CH (Medium)
-
Tags: stack, tree, design, binary search tree, binary tree, iterator
173. Binary Search Tree Iterator - Python Solution
from typing import Optional
from binarytree import Node as TreeNode
from binarytree import build
# BST
class BSTIterator:
def __init__(self, root: Optional[TreeNode]):
self.stack = []
self._leftmost_inorder(root)
def _leftmost_inorder(self, root):
while root:
self.stack.append(root)
root = root.left
def next(self) -> int:
topmost_node = self.stack.pop()
if topmost_node.right:
self._leftmost_inorder(topmost_node.right)
return topmost_node.val
def hasNext(self) -> bool:
return len(self.stack) > 0
root = build([7, 3, 15, None, None, 9, 20])
print(root)
# 7__
# / \
# 3 15
# / \
# 9 20
obj = BSTIterator(root)
print(obj.next()) # 3
print(obj.next()) # 7
print(obj.hasNext()) # True
1586. Binary Search Tree Iterator II¶
-
LeetCode | LeetCode CH (Medium)
-
Tags: stack, tree, design, binary search tree, binary tree, iterator
1586. Binary Search Tree Iterator II - Python Solution
from typing import Optional
from binarytree import Node as TreeNode
from binarytree import build
# BST
class BSTIterator:
def __init__(self, root: Optional[TreeNode]):
self.nodes = self._inorder(root)
self.index = -1
self.size = len(self.nodes)
def _inorder(self, node):
if not node:
return []
return (
self._inorder(node.left) + [node.val] + self._inorder(node.right)
)
def hasNext(self) -> bool:
return self.index < self.size - 1
def next(self) -> int:
self.index += 1
return self.nodes[min(self.index, self.size - 1)]
def hasPrev(self) -> bool:
return self.index > 0
def prev(self) -> int:
self.index -= 1
return self.nodes[max(self.index, 0)]
root = build([7, 3, 15, None, None, 9, 20])
print(root)
# 7__
# / \
# 3 15
# / \
# 9 20
obj = BSTIterator(root)
print(obj.next()) # 3
print(obj.next()) # 7
print(obj.hasNext()) # True
print(obj.prev()) # 3
print(obj.prev()) # None