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Invert Binary Tree⚓︎

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Description⚓︎

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

Solution⚓︎

Recursive Solution⚓︎

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/**
 * 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 nullptr;
        swap(root->left, root->right);
        invertTree(root->left);
        invertTree(root->right);
        return root;
    }
};

Iterative Solution⚓︎

Way 1 (Pre-order)⚓︎

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/**
 * 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) {
        stack<TreeNode*> stk;
        TreeNode *curr = root;

        while (curr || !stk.empty()) {
            if (curr) {
                swap(curr->left, curr->right);
                stk.push(curr);
                curr = curr->left;
            } else {
                auto t = stk.top();
                stk.pop();
                curr = t->right;
            }
        }
        return root;
    }
};

Way 2 (Level-order)⚓︎

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/**
 * 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 nullptr;
        queue<TreeNode*> q;
        q.push(root);

        while (!q.empty()) {
            auto curr = q.front();
            q.pop();

            swap(curr->left, curr->right);
            if (curr->left) q.push(curr->left);
            if (curr->right) q.push(curr->right);
        }
        return root;
    }
};

Explanation:

  1. Check for Empty Tree: If the root is null, the tree is empty, and we return nullptr.
  2. Queue Initialization: We use a queue to keep track of nodes. Initially, it contains only the root.
  3. Tree Traversal: We use a loop to process nodes until the queue is empty. For each node:
    • Swap its left and right children.
    • Enqueue its non-null children for subsequent processing.

Complexity Analysis:

  • Time Complexity: \(O(n)\), where \(n\) is the number of nodes in the tree. Each node is visited exactly once.
  • Space Complexity: \(O(m)\), where m is the maximum number of nodes at any level in the input tree. In the worst case, the space complexity can be \(O(n)\) (consider a perfectly balanced tree).