226_invert_binary_tree.cpp

/*
LeetCode 226 - Invert Binary Tree
Difficulty: Easy

Problem:
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

Time Complexity: O(n)
Space Complexity: O(h) where h is the height of the tree
*/

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

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 nullptr;
        }
        TreeNode* temp = root->left;
        root->left = invertTree(root->right);
        root->right = invertTree(temp);
        return root;
    }
};

vector<int> treeToList(TreeNode* root) {
    vector<int> result;
    if (!root) return result;
    queue<TreeNode*> q;
    q.push(root);
    while (!q.empty()) {
        TreeNode* node = q.front();
        q.pop();
        result.push_back(node->val);
        if (node->left) q.push(node->left);
        if (node->right) q.push(node->right);
    }
    return result;
}

int main() {
    Solution solution;

    TreeNode* root = new TreeNode(4);
    root->left = new TreeNode(2);
    root->right = new TreeNode(7);
    root->left->left = new TreeNode(1);
    root->left->right = new TreeNode(3);
    root->right->left = new TreeNode(6);
    root->right->right = new TreeNode(9);

    TreeNode* inverted = solution.invertTree(root);
    vector<int> result = treeToList(inverted);
    cout << "[";
    for (int i = 0; i < result.size(); i++) {
        cout << result[i];
        if (i < result.size() - 1) cout << ", ";
    }
    cout << "]" << endl; // [4, 7, 2, 9, 6, 3, 1]

    return 0;
}