# 545 Boundary of Binary Tree

## 545. [Boundary of Binary Tree](https://leetcode.com/problems/boundary-of-binary-tree/description/)

## 1. Question

Given a binary tree, return the values of its boundary in **anti-clockwise** direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes.

**Left boundary**is defined as the path from root to the **left-most** node.**Right boundary** is defined as the path from root to the **right-most** node. If the root doesn't have left subtree or right subtree, then the root itself is left boundary or right boundary. Note this definition only applies to the input binary tree, and not applies to any subtrees.

The **left-most** node is defined as a **leaf** node you could reach when you always firstly travel to the left subtree if exists. If not, travel to the right subtree. Repeat until you reach a leaf node.

The **right-most** node is also defined by the same way with left and right exchanged.

**Example 1**

```
Input:

  1
   \
    2
   / \
  3   4


Ouput:

[1, 3, 4, 2]


Explanation:

The root doesn't have left subtree, so the root itself is left boundary.
The leaves are node 3 and 4.
The right boundary are node 1,2,4. Note the anti-clockwise direction means you should output reversed right boundary.
So order them in anti-clockwise without duplicates and we have [1,3,4,2].
```

**Example 2**

```
Input:

    ____1_____
   /          \
  2            3
 / \          / 
4   5        6   
   / \      / \
  7   8    9  10  


Ouput:

[1,2,4,7,8,9,10,6,3]


Explanation:

The left boundary are node 1,2,4. (4 is the left-most node according to definition)
The leaves are node 4,7,8,9,10.
The right boundary are node 1,3,6,10. (10 is the right-most node).
So order them in anti-clockwise without duplicate nodes we have [1,2,4,7,8,9,10,6,3].
```

## 2. Implementation

**(1) Divide into sub problems**

```java
class Solution {
    public List<Integer> boundaryOfBinaryTree(TreeNode root) {
        List<Integer> res = new ArrayList<>();

        if (root == null) {
            return res;
        }
        res.add(root.val);

        getSideBoundary(root.left, res, true);
        getLeaves(root.left, res);
        getLeaves(root.right, res);
        getSideBoundary(root.right, res, false);
        return res;
    }

    public void getSideBoundary(TreeNode node, List<Integer> boundary, boolean isLeft) {
        if (node == null || (node.left == null && node.right == null)) {
            return;
        }

        if (isLeft) {
            boundary.add(node.val);
            if (node.left != null) {
                getSideBoundary(node.left, boundary, isLeft);
            }
            else {
                getSideBoundary(node.right, boundary, isLeft);
            }
        }
        else {
            if (node.right != null) {
                getSideBoundary(node.right, boundary, isLeft);
            }
            else {
                getSideBoundary(node.left, boundary, isLeft);
            }
            boundary.add(node.val);
        }
    }


    public void getLeaves(TreeNode node, List<Integer> boundary) {
        if (node == null) {
            return;
        }

        if (node.left == null && node.right == null) {
            boundary.add(node.val);
            return;
        }

        getLeaves(node.left, boundary);
        getLeaves(node.right, boundary);
    }
}
```

## 3. Time & Space Complexity

**(1) Divide into sub problems:** 时间复杂度: O(n), 空间复杂度: O(n)


---

# Agent Instructions: 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://protegejj.gitbook.io/algorithm-practice/leetcode/tree/545-boundary-of-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.