Today I have done leetcode's October LeetCoding Challenge with cpp.

October LeetCoding Challenge 5

Description

Add One Row to Tree

Given the root of a binary tree and two integers val and depth, add a row of nodes with value val at the given depth depth.

Note that the root node is at depth 1.

The adding rule is:

• Given the integer depth, for each not null tree node cur at the depth depth - 1, create two tree nodes with value val as cur's left subtree root and right subtree root.
• cur's original left subtree should be the left subtree of the new left subtree root.
• cur's original right subtree should be the right subtree of the new right subtree root.
• If depth == 1 that means there is no depth depth - 1 at all, then create a tree node with value val as the new root of the whole original tree, and the original tree is the new root's left subtree.

Example 1:

Input: root = [4,2,6,3,1,5], val = 1, depth = 2
Output: [4,1,1,2,null,null,6,3,1,5]

Example 2:

Input: root = [4,2,null,3,1], val = 1, depth = 3
Output: [4,2,null,1,1,3,null,null,1]

Constraints:

• The number of nodes in the tree is in the range [1, 10^4].
• The depth of the tree is in the range [1, 10^4].
• -100 <= Node.val <= 100
• -105 <= val <= 10^5
• 1 <= depth <= the depth of tree + 1

Solution

class Solution {
public:
  TreeNode* addOneRow(TreeNode* root, int v, int d, bool left = true) {
    if(d == 1) {
      TreeNode *newRoot = new TreeNode(v);
      if(left) newRoot->left = root;
      else newRoot->right = root;
      return newRoot;
    }
    if(!root) return nullptr;
    root->left = addOneRow(root->left, v, d-1, true);
    root->right = addOneRow(root->right, v, d-1, false);
    return root;
  }
};

// Accepted
// 109/109 cases passed (23 ms)
// Your runtime beats 85.7 % of cpp submissions
// Your memory usage beats 62.88 % of cpp submissions (25 MB)