Trees¶

Table of Contents¶

226. Invert Binary Tree¶

LeetCode | LeetCode CH (Easy)
Tags: tree, depth first search, breadth first search, binary tree

226. Invert Binary 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


# Recursive
def invertTreeRecursive(root: Optional[TreeNode]) -> Optional[TreeNode]:
    if not root:
        return root

    root.left, root.right = root.right, root.left

    invertTreeRecursive(root.left)
    invertTreeRecursive(root.right)

    return root


# Iterative
def invertTreeIterative(root: Optional[TreeNode]) -> Optional[TreeNode]:
    if not root:
        return root

    stack = [root]

    while stack:
        node = stack.pop()

        node.left, node.right = node.right, node.left
        if node.left:
            stack.append(node.left)
        if node.right:
            stack.append(node.right)

    return root


root = build([4, 2, 7, 1, 3, 6, 9])
print(root)
#     __4__
#    /     \
#   2       7
#  / \     / \
# 1   3   6   9
invertedRecursive = invertTreeRecursive(root)
print(invertedRecursive)
#     __4__
#    /     \
#   7       2
#  / \     / \
# 9   6   3   1
root = build([4, 2, 7, 1, 3, 6, 9])
invertedIterative = invertTreeIterative(root)
print(invertedIterative)
#     __4__
#    /     \
#   7       2
#  / \     / \
# 9   6   3   1

104. Maximum Depth of Binary Tree¶

LeetCode | LeetCode CH (Easy)
Tags: tree, depth first search, breadth first search, binary tree

104. Maximum Depth of Binary Tree - Python Solution

from collections import deque
from typing import Optional

from binarytree import Node as TreeNode
from binarytree import build


# Recursive
def maxDepthRecursive(root: Optional[TreeNode]) -> int:
    if not root:
        return 0

    left = maxDepthRecursive(root.left)
    right = maxDepthRecursive(root.right)

    return 1 + max(left, right)


# DFS
def maxDepthDFS(root: Optional[TreeNode]) -> int:
    res = 0

    def dfs(node, cnt):
        if node is None:
            return
        cnt += 1
        nonlocal res
        res = max(res, cnt)

        dfs(node.left, cnt)
        dfs(node.right, cnt)

    dfs(root, 0)

    return res


# Iterative
def maxDepthIterative(root: Optional[TreeNode]) -> int:
    if not root:
        return 0

    q = deque([root])
    res = 0

    while q:
        res += 1
        n = len(q)

        for _ in range(n):
            node = q.popleft()
            if node.left:
                q.append(node.left)
            if node.right:
                q.append(node.right)

    return res


root = [1, 2, 2, 3, 4, None, None, None, None, 5]
root = build(root)
print(root)
#     ____1
#    /     \
#   2__     2
#  /   \
# 3     4
#      /
#     5
print(maxDepthRecursive(root))  # 4
print(maxDepthIterative(root))  # 4
print(maxDepthDFS(root))  # 4

100. Same Tree¶

LeetCode | LeetCode CH (Easy)
Tags: tree, depth first search, breadth first search, binary tree

100. Same Tree - Python Solution

from collections import deque
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 isSameTreeRecursive(p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
    if not p and not q:
        return True
    if not p or not q:
        return False
    if p.val != q.val:
        return False

    return isSameTreeRecursive(p.left, q.left) and isSameTreeRecursive(
        p.right, q.right
    )


# 2. Iterative with queue
def isSameTreeIterativeQueue(
    p: Optional[TreeNode], q: Optional[TreeNode]
) -> bool:
    queue = deque([(p, q)])

    while queue:
        p, q = queue.popleft()

        if not p and not q:
            continue
        if not p or not q:
            return False
        if p.val != q.val:
            return False

        queue.append((p.left, q.left))
        queue.append((p.right, q.right))

    return True


# 3. Iterative with stack
def isSameTreeIterativeStack(
    p: Optional[TreeNode], q: Optional[TreeNode]
) -> bool:
    stack = [(p, q)]

    while stack:
        n1, n2 = stack.pop()

        if not n1 and not n2:
            continue
        if not n1 or not n2:
            return False
        if n1.val != n2.val:
            return False

        stack.append((n1.left, n2.left))
        stack.append((n1.right, n2.right))

    return True


p1 = build([1, 2, 3])
q1 = build([1, 2, 3])
p2 = build([1, 2])
q2 = build([1, None, 2])

print(isSameTreeRecursive(p1, q1))  # True
print(isSameTreeRecursive(p2, q2))  # False
print(isSameTreeIterativeQueue(p1, q1))  # True
print(isSameTreeIterativeQueue(p2, q2))  # False
print(isSameTreeIterativeStack(p1, q1))  # True
print(isSameTreeIterativeStack(p2, q2))  # False

572. Subtree of Another Tree¶

LeetCode | LeetCode CH (Easy)
Tags: tree, depth first search, string matching, binary tree, hash function

572. Subtree of Another Tree - Python Solution

from typing import Optional

from binarytree import Node as TreeNode
from binarytree import build


# DFS - Tree
def isSubtree(root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
    def isSameTree(p, q):
        if not p and not q:
            return True
        if not p or not q:
            return False
        if p.val != q.val:
            return False

        return isSameTree(p.left, q.left) and isSameTree(p.right, q.right)

    if not root:
        return False

    return (
        isSameTree(root, subRoot)
        or isSubtree(root.left, subRoot)
        or isSubtree(root.right, subRoot)
    )


# |------------|---------|----------|
# | Approach   | Time    | Space    |
# |------------|---------|----------|
# | DFS        | O(n * m)| O(n)     |
# |------------|---------|----------|


root = build([3, 4, 5, 1, 2, None, None, None, None, 0])
subRoot = build([4, 1, 2])
print(root)
#     ____3
#    /     \
#   4__     5
#  /   \
# 1     2
#      /
#     0
print(subRoot)
#   4
#  / \
# 1   2
print(isSubtree(root, subRoot))  # False

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

102. Binary Tree Level Order Traversal¶

LeetCode | LeetCode CH (Medium)
Tags: tree, breadth first search, binary tree

102. Binary Tree Level Order Traversal - Python Solution

from collections import deque
from typing import List, Optional

from binarytree import Node as TreeNode
from binarytree import build


def levelOrder(root: Optional[TreeNode]) -> List[List[int]]:
    if not root:
        return []

    q = deque([root])
    res = []

    while q:
        level = []
        size = len(q)

        for _ in range(size):
            cur = q.popleft()
            level.append(cur.val)

            if cur.left:
                q.append(cur.left)
            if cur.right:
                q.append(cur.right)

        res.append(level)

    return res


tree = build([3, 9, 20, None, None, 15, 7])
print(tree)
#   3___
#  /    \
# 9     _20
#      /   \
#     15    7
print(levelOrder(tree))  # [[3], [9, 20], [15, 7]]

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;
}

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

105. Construct Binary Tree from Preorder and Inorder Traversal¶

LeetCode | LeetCode CH (Medium)
Tags: array, hash table, divide and conquer, tree, binary tree

105. Construct Binary Tree from Preorder and Inorder Traversal - Python Solution

from typing import List, Optional

from helper import TreeNode


def buildTree(preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
    """
    preorder  root left right  1  2  3
    inorder   left root right  4  5  6
    """
    if len(preorder) == 0:
        return None

    root_val = preorder[0]  # 1
    root = TreeNode(root_val)

    separator_idx = inorder.index(root_val)  # 5

    left_inorder = inorder[:separator_idx]  # 4
    right_inorder = inorder[separator_idx + 1 :]  # 6

    left_preorder = preorder[1 : 1 + len(left_inorder)]  # 2
    right_preorder = preorder[1 + len(left_inorder) :]  # 3

    root.left = buildTree(left_preorder, left_inorder)
    root.right = buildTree(right_preorder, right_inorder)

    return root


preorder = [3, 9, 20, 15, 7]
inorder = [9, 3, 15, 20, 7]
root = buildTree(preorder, inorder)
print(root)
#     3
#    / \
#   9  20
#     /  \
#    15   7

105. Construct Binary Tree from Preorder and Inorder Traversal - C++ Solution

#include <iostream>
#include <unordered_map>
#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:
    vector<int> preorder;
    unordered_map<int, int> inorderMap;

    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        this->preorder = preorder;
        for (size_t i = 0; i < inorder.size(); i++) {
            inorderMap[inorder[i]] = i;
        }
        return buildSubtree(0, 0, inorder.size() - 1);
    }

   private:
    TreeNode* buildSubtree(int rootIndex, int left, int right) {
        if (left > right) return nullptr;

        TreeNode* root = new TreeNode(preorder[rootIndex]);
        int inorderIndex = inorderMap[preorder[rootIndex]];

        root->left = buildSubtree(rootIndex + 1, left, inorderIndex - 1);
        root->right = buildSubtree(rootIndex + (inorderIndex - left + 1),
                                   inorderIndex + 1, right);

        return root;
    }
};

int main() {
    vector<int> preorder = {3, 9, 20, 15, 7};
    vector<int> inorder = {9, 3, 15, 20, 7};
    Solution solution;
    TreeNode* root = solution.buildTree(preorder, inorder);
    cout << root->val << endl;                // 3
    cout << root->left->val << endl;          // 9
    cout << root->right->val << endl;         // 20
    cout << root->right->left->val << endl;   // 15
    cout << root->right->right->val << endl;  // 7
    return 0;
}

124. Binary Tree Maximum Path Sum¶

LeetCode | LeetCode CH (Hard)
Tags: dynamic programming, tree, depth first search, binary tree

124. Binary Tree Maximum Path Sum - Python Solution

from typing import Optional

from binarytree import Node as TreeNode
from binarytree import build


def maxPathSum(root: Optional[TreeNode]) -> int:
    res = float("-inf")

    def dfs(node):
        if not node:
            return 0

        leftMax = max(dfs(node.left), 0)
        rightMax = max(dfs(node.right), 0)

        cur = node.val + leftMax + rightMax
        nonlocal res
        res = max(res, cur)

        return node.val + max(leftMax, rightMax)

    dfs(root)

    return res


root = build([-10, 9, 20, None, None, 15, 7])
print(root)
#   -10___
#  /      \
# 9       _20
#        /   \
#       15    7
print(maxPathSum(root))  # 42

298. Binary Tree Longest Consecutive Sequence¶

LeetCode | LeetCode CH (Medium)
Tags: tree, depth first search, binary tree

298. Binary Tree Longest Consecutive Sequence - Python Solution

from typing import Optional

from binarytree import Node as TreeNode
from binarytree import build


# Binary Tree
def longestConsecutive(root: Optional[TreeNode]) -> int:
    res = 0

    def dfs(node):
        if not node:
            return 0

        left, right = dfs(node.left), dfs(node.right)
        cur = 1
        if node.left and node.left.val == (node.val + 1):
            cur = max(cur, left + 1)
        if node.right and node.right.val == (node.val + 1):
            cur = max(cur, right + 1)

        nonlocal res
        res = max(res, cur)
        return cur

    dfs(root)

    return res


if __name__ == "__main__":
    root = build([1, 3, 2, 4, None, None, None, 5])
    print(root)
    #       1
    #      / \
    #     3   2
    #    /
    #   4
    #  /
    # 5
    print(longestConsecutive(root))  # 3