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# 226 Invert Binary Tree

## 226. Invert Binary Tree

## 1. Question

Invert a binary tree.

```
     4
   /   \
  2     7
 / \   / \
1   3 6   9
```

to

```
     4
   /   \
  7     2
 / \   / \
9   6 3   1
```

## 2. Implementation

**(1) DFS**

```java
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return null;
        }

        TreeNode left = root.left;
        TreeNode right = root.right;
        root.left = invertTree(right);
        root.right = invertTree(left);
        return root;
    }
}
```

**（2） BFS**

```java
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return null;
        }

        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);

        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode curNode = queue.remove();
                TreeNode temp = curNode.left;
                curNode.left = curNode.right;
                curNode.right = temp;

                if (curNode.left != null) {
                    queue.add(curNode.left);
                }

                if (curNode.right != null) {
                    queue.add(curNode.right);
                }
            }
        }
        return root;
    }
}
```

## 3. Time & Space Complexity

DFS: 时间复杂度: O(n), 空间复杂度O(n), 平均复杂度是O(logn), 当树是平衡的时候

BFS: 时间复杂度: O(n), 空间复杂度O(w), w为一行中node最多的个数