Posts tagged as “tree”

花花酱 LeetCode 103. Binary Tree Zigzag Level Order Traversal

By zxi on August 20, 2019

Given a binary tree, return the zigzag level order traversal of its nodes’ values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree [3,9,20,null,null,15,7],

return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]

Solution 1: DFS

in order traversal using DFS and reverse the result of even levels.

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua, 0ms, 15.1 MB

class Solution {

public:

vector<vector<int> > zigzagLevelOrder(TreeNode *root) {

vector<vector<int>> ans;

function<void(TreeNode*, int)> dfs = [&](TreeNode* r, int d) {

if (!r) return;

while (ans.size() <= d) ans.push_back({});

ans[d].push_back(r->val);

dfs(r->right, d + 1);

dfs(r->left, d + 1);

};

dfs(root, 0);

for (int i = 0; i < ans.size(); i += 2)

reverse(begin(ans[i]), end(ans[i]));

return ans;

}

};

Solution 2: BFS

Expend/append in order for even levels and doing that in reverse order for odd levels.

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

vector<vector<int>> zigzagLevelOrder(TreeNode *root) {

if (!root) return {};

vector<vector<int>> ans;

deque<TreeNode*> q;

q.push_back(root);

int d = 0;

while (q.size()) {

ans.push_back({});

auto cur = &ans.back();

deque<TreeNode*> next;

while (q.size()) {

if (d % 2) {

TreeNode* node = q.back();

q.pop_back();

cur->push_back(node->val);

if (node->right) next.push_front(node->right);

if (node->left) next.push_front(node->left);

} else {

TreeNode* node = q.front();

q.pop_front();

cur->push_back(node->val);

if (node->left) next.push_back(node->left);

if (node->right) next.push_back(node->right);

}

++d;

q.swap(next);

}

return ans;

}

};

花花酱 LeetCode 1161. Maximum Level Sum of a Binary Tree

By zxi on August 17, 2019

Given the root of a binary tree, the level of its root is 1, the level of its children is 2, and so on.

Return the smallest level X such that the sum of all the values of nodes at level X is maximal.

Example 1:

Input: [1,7,0,7,-8,null,null]
Output: 2
Explanation: 
Level 1 sum = 1.
Level 2 sum = 7 + 0 = 7.
Level 3 sum = 7 + -8 = -1.
So we return the level with the maximum sum which is level 2.

Note:

The number of nodes in the given tree is between 1 and 10^4.
-10^5 <= node.val <= 10^5

Solution: HashTable

Use a hash table / array to store the level sum.

Time complexity: O(n)
Space complexity: O(h)

C++

// Author: Huahua, 232 ms, 70.2 ms

class Solution {

public:

int maxLevelSum(TreeNode* root) {

vector<int> sums;

function<void(TreeNode*, int)> preorder = [&](TreeNode* n, int d) {

if (!n) return;

while (sums.size() <= d) sums.push_back(0);

sums[d] += n->val;

preorder(n->left, d + 1);

preorder(n->right, d + 1);

};

preorder(root, 1);

int max_sum = sums[1];

int ans = 1;

for (int i = 2; i < sums.size(); ++i)

if (sums[i] > max_sum) {

max_sum = sums[i];

ans = i;

}

return ans;

}

};

花花酱 LeetCode 1110. Delete Nodes And Return Forest

By zxi on August 8, 2019

Given the root of a binary tree, each node in the tree has a distinct value.

After deleting all nodes with a value in to_delete, we are left with a forest (a disjoint union of trees).

Return the roots of the trees in the remaining forest. You may return the result in any order.

Example 1:

Input: root = [1,2,3,4,5,6,7], to_delete = [3,5]
Output: [[1,2,null,4],[6],[7]]

Constraints:

The number of nodes in the given tree is at most 1000.
Each node has a distinct value between 1 and 1000.
to_delete.length <= 1000
to_delete contains distinct values between 1 and 1000.

Solution: Recursion / Post-order traversal

Recursively delete nodes on left subtree and right subtree and return the trimmed tree.
if the current node needs to be deleted, then its non-null children will be added to output array.

Time complexity: O(n)
Space complexity: O(|d| + h)

C++

// Author: Huahua

class Solution {

public:

vector<TreeNode*> delNodes(TreeNode* root, vector<int>& to_delete) {

vector<TreeNode*> ans;

unordered_set<int> s{begin(to_delete), end(to_delete)};

function<TreeNode*(TreeNode*)> del = [&](TreeNode* n) -> TreeNode* {

if (!n) return nullptr;

if (n->left) n->left = del(n->left);

if (n->right) n->right = del(n->right);

if (!s.count(n->val)) return n;

if (n->left) ans.push_back(n->left);

if (n->right) ans.push_back(n->right);

return nullptr;

};

TreeNode* r = del(root);

if (r) ans.push_back(r);

return ans;

}

};

Python3

# Author: Huahua

class Solution:

def delNodes(self, root: TreeNode, to_delete: List[int]) -> List[TreeNode]:

ans = []

ds = set(to_delete)

def process(n):

if not n: return None

n.left, n.right = process(n.left), process(n.right)

if n.val not in ds: return n

if n.left: ans.append(n.left)

if n.right: ans.append(n.right)

return None

root = process(root)

if root: ans.append(root)

return ans

花花酱 LeetCode 1145. Binary Tree Coloring Game

By zxi on August 4, 2019

Two players play a turn based game on a binary tree. We are given the root of this binary tree, and the number of nodes n in the tree. n is odd, and each node has a distinct value from 1 to n.

Initially, the first player names a value x with 1 <= x <= n, and the second player names a value y with 1 <= y <= n and y != x. The first player colors the node with value x red, and the second player colors the node with value yblue.

Then, the players take turns starting with the first player. In each turn, that player chooses a node of their color (red if player 1, blue if player 2) and colors an uncolored neighbor of the chosen node (either the left child, right child, or parent of the chosen node.)

If (and only if) a player cannot choose such a node in this way, they must pass their turn. If both players pass their turn, the game ends, and the winner is the player that colored more nodes.

You are the second player. If it is possible to choose such a y to ensure you win the game, return true. If it is not possible, return false.

Example 1:

Input: root = [1,2,3,4,5,6,7,8,9,10,11], n = 11, x = 3
Output: true
Explanation: The second player can choose the node with value 2.

Constraints:

root is the root of a binary tree with n nodes and distinct node values from 1 to n.
n is odd.
1 <= x <= n <= 100

Solution: Count size of red’s subtrees

There are two situations that blue can win.
1. one of the red’s subtree has more than n>>1 nodes. Blue colorize the root of the larger subtree.
2. red and its children has size less or equal to n>>1. Blue colorize red’s parent.

Time complexity: O(n)
Space complexity: O(h)

C++

// Author: Huahua

class Solution {

public:

bool btreeGameWinningMove(TreeNode* root, int n, int x) {

int red_left;

int red_right;

function<int(TreeNode*)> count = [&](TreeNode* node){

if (!node) return 0;

int l = count(node->left);

int r = count(node->right);

if (root->val == x) {

red_left = l;

red_right = r;

}

return 1 + l + r;

};

count(root);

// Color red's parent.

if (1 + red_left + red_right <= n / 2) return true;

// Color the child that has the larger subtree.

return max(red_left, red_right) > n / 2;

}

};

花花酱 LeetCode 226. Invert Binary Tree

By zxi on July 22, 2019

Invert a binary tree.

Example:

Input:

     4
   /   \
  2     7
 / \   / \
1   3 6   9

Output:

     4
   /   \
  7     2
 / \   / \
9   6 3   1

Trivia:
This problem was inspired by this original tweet by Max Howell:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so f*** off.

Solution: Recursion

Recursive invert the left and right subtrees and swap them.

Time complexity: O(n)
Space complexity: O(h)

C++

// Author: Huahua

class Solution {

public:

TreeNode* invertTree(TreeNode* root) {

if (!root) return nullptr;

TreeNode* tmp = root->right;

root->right = invertTree(root->left);

root->left = invertTree(tmp);

return root;

}

};

Python3

# Author: Huahua

class Solution:

def invertTree(self, root: TreeNode) -> TreeNode:

if not root: return None

root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)

return root

Posts tagged as “tree”

花花酱 LeetCode 103. Binary Tree Zigzag Level Order Traversal

Solution 1: DFS

C++

Solution 2: BFS

C++

Related Problems

花花酱 LeetCode 1161. Maximum Level Sum of a Binary Tree

Solution: HashTable

C++

花花酱 LeetCode 1110. Delete Nodes And Return Forest

Solution: Recursion / Post-order traversal

C++

Python3

花花酱 LeetCode 1145. Binary Tree Coloring Game

C++

花花酱 LeetCode 226. Invert Binary Tree

Solution: Recursion

C++

Python3