recursion Archives - Page 5 of 14

花花酱 LeetCode 257. Binary Tree Paths

By zxi on January 29, 2020

Given a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

Example:

Input:

   1
 /   \
2     3
 \
  5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

Solution: Recursion

Time complexity: O(n)
Space complexity: O(n) / output can be O(n^2)

C++

// Author: Huahua

class Solution {

public:

vector<string> binaryTreePaths(TreeNode* root) {

vector<string> ans;

string s;

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

if (!node) return;

s += (l > 0 ? "->" : "") + to_string(node->val);

if (!node->left && !node->right)

ans.push_back(s);

preorder(node->left, s.size());

preorder(node->right, s.size());

while (s.size() != l) s.pop_back();

};

preorder(root, 0);

return ans;

}

};

花花酱 LeetCode 107. Binary Tree Level Order Traversal II

By zxi on January 29, 2020

Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).

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

return its bottom-up level order traversal as:

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

Solution: Recursion

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

C++

// Author: Huahua

class Solution {

public:

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

vector<vector<int>> ans;

levelOrderBottom(root, 0, ans);

reverse(ans.begin(), ans.end());

return ans;

}

private:

void levelOrderBottom(TreeNode *root, int depth, vector<vector<int>>& ans) {

if (!root) return;

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

ans[depth].push_back(root->val);

levelOrderBottom(root->left, depth + 1, ans);

levelOrderBottom(root->right, depth + 1, ans);

}

};

花花酱 LeetCode 144. Binary Tree Preorder Traversal

By zxi on January 29, 2020

Given a binary tree, return the preorder traversal of its nodes’ values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,2,3]

Follow up: Recursive solution is trivial, could you do it iteratively?

Solution 1: Recursion

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

C++

// Author: Huahua

class Solution {

public:

vector<int> preorderTraversal(TreeNode* root) {

vector<int> ans;

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

if (!n) return;

ans.push_back(n->val);

preorder(n->left);

preorder(n->right);

};

preorder(root);

return ans;

}

};

Solution 2: Stack

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

C++

// Author: Huahua

class Solution {

public:

vector<int> preorderTraversal(TreeNode* root) {

vector<int> ans;

stack<TreeNode*> s;

if (root) s.push(root);

while (!s.empty()) {

TreeNode* n = s.top();

ans.push_back(n->val);

s.pop();

if (n->right) s.push(n->right);

if (n->left) s.push(n->left);

}

return ans;

}

花花酱 LeetCode 1325. Delete Leaves With a Given Value

By zxi on January 20, 2020

Given a binary tree root and an integer target, delete all the leaf nodes with value target.

Note that once you delete a leaf node with value target, if it’s parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you can’t).

Example 1:

Input: root = [1,2,3,2,null,2,4], target = 2
Output: [1,null,3,null,4]
Explanation: Leaf nodes in green with value (target = 2) are removed (Picture in left). 
After removing, new nodes become leaf nodes with value (target = 2) (Picture in center).

Example 2:

Input: root = [1,3,3,3,2], target = 3
Output: [1,3,null,null,2]

Example 3:

Input: root = [1,2,null,2,null,2], target = 2
Output: [1]
Explanation: Leaf nodes in green with value (target = 2) are removed at each step.

Example 4:

Input: root = [1,1,1], target = 1
Output: []

Example 5:

Input: root = [1,2,3], target = 1
Output: [1,2,3]

Constraints:

1 <= target <= 1000
Each tree has at most 3000 nodes.
Each node’s value is between [1, 1000].

Solution: Recursion

Post-order traversal

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

C++

// Author: Huahua

class Solution {

public:

TreeNode* removeLeafNodes(TreeNode* root, int target) {

if (!root) return nullptr;

root->left = removeLeafNodes(root->left, target);

root->right = removeLeafNodes(root->right, target);

return root->left || root->right || root->val != target

? root : nullptr;

}

};

花花酱 LeetCode 1315. Sum of Nodes with Even-Valued Grandparent

By zxi on January 11, 2020

Given a binary tree, return the sum of values of nodes with even-valued grandparent. (A grandparent of a node is the parent of its parent, if it exists.)

If there are no nodes with an even-valued grandparent, return 0.

Example 1:

Input: root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5]
Output: 18
Explanation: The red nodes are the nodes with even-value grandparent while the blue nodes are the even-value grandparents.

Constraints:

The number of nodes in the tree is between 1 and 10^4.
The value of nodes is between 1 and 100.

Solution: Recursion

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

C++

// Author: Huahua

class Solution {

public:

int sumEvenGrandparent(TreeNode* root, int p = 1, int gp = 1) {

if (!root) return 0;

return sumEvenGrandparent(root->left, root->val, p)

+ sumEvenGrandparent(root->right, root->val, p)

+ (gp & 1 ? 0 : root->val);

}

};

Posts tagged as “recursion”

花花酱 LeetCode 257. Binary Tree Paths

Solution: Recursion

C++

花花酱 LeetCode 107. Binary Tree Level Order Traversal II

Solution: Recursion

C++

Related Problems

花花酱 LeetCode 144. Binary Tree Preorder Traversal

Solution 1: Recursion

C++

Solution 2: Stack

C++

花花酱 LeetCode 1325. Delete Leaves With a Given Value

Solution: Recursion

C++

花花酱 LeetCode 1315. Sum of Nodes with Even-Valued Grandparent

C++