226. Invert Binary Tree

Solution - Recursion

/**
 * Definition for a binary tree node.
 * 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* invertTree(TreeNode* root)
    {
        if (!root) return NULL;
        
        swap(root->left, root->right);
        
        invertTree(root->left);
        invertTree(root->right);
        return root;
    }
};

• T:
  $O(N)$
• S:
  $O(H)$

Solution - Iteration

/**
 * Definition for a binary tree node.
 * 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* invertTree(TreeNode* root)
    {
        if (!root) return NULL;
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty())
        {
            TreeNode* node = q.front(); q.pop();

            swap(node->left, node->right);

            if(node->left) q.push(node->left);
            
            if(node->right) q.push(node->right);
        }
        return root;
    }
};

• T:
  $O(N)$
• S:
  $O(N)$