> For the complete documentation index, see [llms.txt](https://grind75-notes.gitbook.io/notes/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://grind75-notes.gitbook.io/notes/week-1/invert-binary-tree.md).

# Invert Binary Tree

{% embed url="<https://leetcode.com/problems/invert-binary-tree/>" %}

### Problem

> Given the `root` of a binary tree, invert the tree, and return *its root*.
>
> &#x20;
>
> **Example 1:**
>
> ![](https://assets.leetcode.com/uploads/2021/03/14/invert1-tree.jpg)
>
> <pre><code>Input: root = [4,2,7,1,3,6,9]
> <strong>Output:
> </strong> [4,7,2,9,6,3,1]
> </code></pre>
>
> **Example 2:**
>
> ![](https://assets.leetcode.com/uploads/2021/03/14/invert2-tree.jpg)
>
> <pre><code>Input: root = [2,1,3]
> <strong>Output:
> </strong> [2,3,1]
> </code></pre>
>
> **Example 3:**
>
> <pre><code>Input: root = []
> <strong>Output:
> </strong> []
> </code></pre>

### Pseudocode

{% code overflow="wrap" %}

```
inverting a tree involves swapping the left and right child nodes
generally
let temp = node.left
node.left = node.right
node.right = temp

in the example for the tree with 3 node height, the left most node is inverted to be the right most node.

two methods to traverse a binary tree. dfs, bfs.

dfs solution
swap left and right nodes in post traversal

bfs solution

```

{% endcode %}

### Solution

```javascript
// dfs
var invertTree = function (root) {
  function walk(node) {
    // base condition
    if (!node) {
      return;
    }

    // pre
    // console.log(node.val)
    // recurse
    walk(node.left);
    walk(node.right);
    // post
    let temp = node.left;
    node.left = node.right;
    node.right = temp;
  }

  walk(root);

  return root;
};

//bfs
var invertTree = function(root) {
    if(!root) {
        return null;
    }

    let q = [root];

    while(q.length) {
        const node = q.pop()

        if(node.left) q.push(node.left);
        if(node.right) q.push(node.right);

        let temp = node.left;
        node.left = node.right;
        node.right = temp;
    }

    return root
};

```

### Time and Space Complexity

#### Time

* While traversal/searching of BT is usually O(height), inversion requires complete traversal of all inputs, regardless of bfs or dfs. O(N)
* Total - O(N)

#### Space

* Returning the same node length O(N)
* Total - O(N)


---

# Agent Instructions
This documentation is published with GitBook. GitBook is the documentation platform designed so that both humans and AI agents can read, navigate, and reason over technical content effectively. Learn more at gitbook.com.

## Querying This Documentation
If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://grind75-notes.gitbook.io/notes/week-1/invert-binary-tree.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.