226. Invert Binary Tree

Recursion

/*
 * @lc app=leetcode id=226 lang=cpp
 *
 * [226] Invert Binary Tree
 *
 * https://leetcode.com/problems/invert-binary-tree/description/
 *
 * algorithms
 * Easy (78.10%)
 * Likes:    14546
 * Dislikes: 241
 * Total Accepted:    2.6M
 * Total Submissions: 3.2M
 * Testcase Example:  '[4,2,7,1,3,6,9]'
 *
 * Given the root of a binary tree, invert the tree, and return its root.
 *
 *
 * Example 1:
 *
 *
 * Input: root = [4,2,7,1,3,6,9]
 * Output: [4,7,2,9,6,3,1]
 *
 *
 * Example 2:
 *
 *
 * Input: root = [2,1,3]
 * Output: [2,3,1]
 *
 *
 * Example 3:
 *
 *
 * Input: root = []
 * Output: []
 *
 *
 *
 * Constraints:
 *
 *
 * The number of nodes in the tree is in the range [0, 100].
 * -100 <= Node.val <= 100
 *
 *
 */

// @lc code=start
/**
 * 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;
    }
};
// @lc code=end

• T: $O(N)$

• S: $O(H)$

• 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: $O(N)$