This approach employs recursion to invert the binary tree. The basic idea is to traverse the tree and swap the left and right children of every node recursively. We recursively call the function on the left and right subtrees until we reach the base case, which is when the node is null.
Time Complexity: O(n) where n is the number of nodes in the tree, because each node is visited once.
Space Complexity: O(h) where h is the height of the tree, for the recursion stack.
1#include <stdio.h>
2#include <stdlib.h>
3
4struct TreeNode {
5 int val;
6 struct TreeNode *left;
7 struct TreeNode *right;
8};
9
10struct TreeNode* invertTree(struct TreeNode* root) {
11 if (root == NULL) return NULL;
12 struct TreeNode* left = invertTree(root->left);
13 struct TreeNode* right = invertTree(root->right);
14 root->left = right;
15 root->right = left;
16 return root;
17}This C solution defines a struct for the tree node, and then uses a recursive function invertTree that swaps the left and right children of each node starting from the root. The recursion continues until all nodes have been swapped.
The iterative approach involves using a queue or stack to traverse the tree level by level (or depth by depth) while inverting the child nodes by putting the left child in place of the right child and vice versa for each node. This is effectively a breadth-first traversal that swaps children iteratively.
Time Complexity: O(n) where n is the number of nodes in the tree.
Space Complexity: O(n) due to the queue data structure.
1import java.util.LinkedList;
2import java.util.Queue;
3
4class TreeNode {
5 int val;
6 TreeNode left;
7 TreeNode right;
8 TreeNode(int x) { val = x; }
9}
10
11class Solution {
12 public TreeNode invertTree(TreeNode root) {
13 if (root == null) return null;
14 Queue<TreeNode> queue = new LinkedList<>();
15 queue.add(root);
16 while (!queue.isEmpty()) {
17 TreeNode current = queue.poll();
18 TreeNode temp = current.left;
19 current.left = current.right;
20 current.right = temp;
21 if (current.left != null) queue.add(current.left);
22 if (current.right != null) queue.add(current.right);
23 }
24 return root;
25 }
26}The Java solution applies an iterative approach using the LinkedList as a queue. Nodes are swapped at each level using a queue to keep track of the current level of nodes being processed.