0226 - Invert Binary Tree

0226 - Invert Binary Tree

Given the root of a binary tree, invert the tree, and return its root.

Examples

Input: root = [4,2,7,1,3,6,9] Output: [4,7,2,9,6,3,1] Input: root = [2,1,3] Output: [2,3,1]

Input: root = [] Output: []

Constraints

The number of nodes in the tree is in the range [0, 100]. -100 <= Node.val <= 100

Java Solution

Recursive

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root == null) return null;
        
        TreeNode right = invertTree(root.right);
        TreeNode left = invertTree(root.left);
        root.left = right;
        root.right = left;
        
        return root;
    }
}

Iterative (Breadth first search)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        while (!queue.isEmpty()) {
            TreeNode current = queue.poll();
            TreeNode temp = current.left;
            current.left = current.right;
            current.right = temp;
            if (current.left != null) queue.add(current.left);
            if (current.right != null) queue.add(current.right);
        }
        return root;
    }
}