Solution 1DFS 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;
 *     }
 * }
 */

public TreeNode invertTree(TreeNode root) {
    if (root == null) {
        return null;
    }
    TreeNode temp = root.left;
    root.left = invertTree(root.right);
    root.right = invertTree(temp);
    return root;
}
Leet Code/java.java · L1005–1030
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var invertTree = function(root) {
    if (!root) return null;
    const temp = root.left;
    root.left = invertTree(root.right);
    root.right = invertTree(temp);
    return root;
};
Leet Code/javascript.js · L48–67
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */
function invertTree(root: TreeNode | null): TreeNode | null {
    if (!root) return null;
    const temp: TreeNode | null = root.left;
    root.left = invertTree(root.right);
    root.right = invertTree(temp);
    return root;
};
Leet Code/typesript.ts · L62–82