Reverse a binary tree in O(1) in C/C++.

struct NormalNode {
  int value;
  struct NormalNode *left;
  struct NormalNode *right;
};

struct ReversedNode {
  int value;
  struct ReversedNode *right;
  struct ReversedNode *left;
};

struct ReversedNode *reverseTree(struct NormalNode *root) {
  return (struct ReversedNode *)root;
}

There are a couple interesting parts to this question. First, since your language is Java, you're most likely to have a generic Node class, something like this:

class Node<T> {
    private final T data;
    private final Node left;
    private final Node right;
    public Node<T>(final T data, final Node left, final Node right) {
        this.data  = data;
        this.left  = left;
        this.right = right;
    }
    ....
}

Secondly, reversing, sometimes called inverting, can be done either by mutating the left and right fields of the node, or by creating a new node just like the original but with its left and right children "reversed." The former approach is shown in another answer, while the second approach is shown here:

class Node<T> {
    // See fields and constructor above...

    public Node<T> reverse() {
        Node<T> newLeftSubtree = right == null ? null : right.reverse();
        Node<T> newRightSubtree = left == null ? null : left.reverse();
        return Node<T>(data, newLeftSubtree, newRightSubtree); 
    }
}

The idea of not mutating a data structure is one of the ideas behind persistent data structures, which are pretty interesting.

You can recursively swap the left and right nodes as below;

// helper method
private static void reverseTree(TreeNode<Integer> root) {
    reverseTreeNode(root);
}

private static void reverseTreeNode(TreeNode<Integer> node) {
    TreeNode<Integer> temp = node.left;
    node.left   = node.right;
    node.right  = temp;

    if(node.left != null)
        reverseTreeNode(node.left);

    if(node.right != null)
        reverseTreeNode(node.right);
}

Demonstration Code for Java

import java.util.LinkedList;
import java.util.Queue;

public class InvertBinaryTreeDemo {

    public static void main(String[] args) {

        // root node
        TreeNode<Integer> root  = new TreeNode<>(6);

        // children of root
        root.left               = new TreeNode<Integer>(3);
        root.right              = new TreeNode<Integer>(4);

        // grand left children of root
        root.left.left          = new TreeNode<Integer>(7);
        root.left.right         = new TreeNode<Integer>(3);

        // grand right childrend of root
        root.right.left         = new TreeNode<Integer>(8);
        root.right.right        = new TreeNode<Integer>(1);

        System.out.println("Before invert");
        traverseTree(root);

        reverseTree(root);

        System.out.println("\nAfter invert");
        traverseTree(root);
    }

    // helper method
    private static void reverseTree(TreeNode<Integer> root) {
        reverseTreeNode(root);
    }

    private static void reverseTreeNode(TreeNode<Integer> node) {
        TreeNode<Integer> temp = node.left;
        node.left   = node.right;
        node.right  = temp;

        if(node.left != null)
            reverseTreeNode(node.left);

        if(node.right != null)
            reverseTreeNode(node.right);
    }

    // helper method for traverse
    private static void traverseTree(TreeNode<Integer> root) {
        Queue<Integer> leftChildren     = new LinkedList<>();
        Queue<Integer> rightChildren    = new LinkedList<>();

        traverseTreeNode(root, leftChildren, rightChildren);

        System.out.println("Tree;\n*****");

        System.out.printf("%3d\n", root.value);

        int count = 0;
        int div = 0;
        while(!(leftChildren.isEmpty() && rightChildren.isEmpty())) {
            System.out.printf("%3d\t%3d\t", leftChildren.poll(), rightChildren.poll());
            count += 2;
            div++;
            if( (double)count == (Math.pow(2, div))) {
                System.out.println();
                count = 0;
            }
        }

        System.out.println();
    }

    private static void traverseTreeNode(TreeNode<Integer> node, Queue<Integer> leftChildren, Queue<Integer> rightChildren) {
        if(node.left != null)
            leftChildren.offer(node.left.value);

        if(node.right != null)
            rightChildren.offer(node.right.value);

        if(node.left != null) {
            traverseTreeNode(node.left, leftChildren, rightChildren);
        }

        if(node.right != null) {
            traverseTreeNode(node.right, leftChildren, rightChildren);
        }
    }

    private static class TreeNode<E extends Comparable<E>> {

        protected E value;
        protected TreeNode<E> left;
        protected TreeNode<E> right;

        public TreeNode(E value) {
            this.value = value;
            this.left = null;
            this.right = null;
        }

    }

}

Output

Before invert
Tree;
*****
  6
  3   4 
  7   3   8   1 

After invert
Tree;
*****
  6
  4   3 
  1   8   3   7 

The recursion function can be very simple as shown below:

public Node flipTree(Node node) {
    if(node == null) return null;

    Node left = flipTree(node.left);
    Node right = flipTree(node.right);

    node.left = right;
    node.right = left;

    return node;
}

I've seen most of the answers aren't focussing on null pointer issues.

public static Node invertBinaryTree(Node node) {

    if(node != null) {
        Node temp = node.getLeftChild();

        node.setLeftChild(node.getRightChild());
        node.setRigthChild(temp);

        if(node.left!=null) { 
            invertBinaryTree(node.getLeftChild());
        }
        if(node.right !=null) {
            invertBinaryTree(node.getRightChild());
        }
    }

    return node;
}

In the code above we are making recursive calls only if the left/right child of the root node isn't null. Its one of the fastest approach!