Example

1         1 / \       / \2   3  => 3   2   /       \  4         4

Solution

Recursion:

public class Solution {    public void invertBinaryTree(TreeNode root) {        if (root == null) return;        TreeNode temp = root.left;        root.left = root.right;        root.right = temp;        invertBinaryTree(root.left);        invertBinaryTree(root.right);        return;    }}

Queue/linkedlist:

用queue的方法要熟练掌握。

public class Solution {    public void invertBinaryTree(TreeNode root) {        if (root == null) return;        Queue
    
      queue = new LinkedList
     
      ();        queue.offer(root);        while (!queue.isEmpty()) {            TreeNode node = queue.poll();            TreeNode temp = node.left;            node.left = node.right;            node.right = temp;            if (node.left != null) {                queue.offer(node.left);            }            if (node.right != null) {                queue.offer(node.right);            }        }        return;    }}