---
tags: leetcode
---

# [545. Boundary of Binary Tree](https://leetcode.com/problems/boundary-of-binary-tree/)
---
# My Solution
## The Key Idea for Solving This Coding Question
## C++ Code

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */

class Solution {
public:
    vector<int> boundaryOfBinaryTree(TreeNode *root) {
        vector<int> boundary;
        
        boundary.push_back(root->val);
        
        if (root->left != nullptr) {
            leftBoundary(root->left, boundary);
        }
        
        if (root->left != nullptr || root->right != nullptr) {
            bottomBoundary(root, boundary);
        }
        
        if (root->right != nullptr) {
            rightBoundary(root->right, boundary);
        }
        
        return boundary;
    }
    
private:
    void leftBoundary(TreeNode *root, vector<int> &boundary) {
        if (root == nullptr || 
            (root->left == nullptr && root->right == nullptr)) {
            return;
        }
        
        boundary.push_back(root->val);
        
        if (root->left != nullptr) {
            leftBoundary(root->left, boundary);
            return;
        }
        leftBoundary(root->right, boundary);
    }
    
    void bottomBoundary(TreeNode *root, vector<int> &boundary) {
        if (root == nullptr) {
            return;
        }
        
        if (root->left == nullptr && root->right == nullptr) {
            boundary.push_back(root->val);
        }
        bottomBoundary(root->left, boundary);
        bottomBoundary(root->right, boundary);
    }
    
    void rightBoundary(TreeNode *root, vector<int> &boundary) {
        if (root == nullptr || 
            (root->left == nullptr && root->right == nullptr)) {
            return;
        }
        
        if (root->right != nullptr) {
            rightBoundary(root->right, boundary);
            boundary.push_back(root->val);
            return;
        }
        rightBoundary(root->left, boundary);
        boundary.push_back(root->val);
    }
};

## Time Complexity
$O(n)$
$n$ is the number of nodes in the binary tree referred by `root`.
## Space Complexity
$O(H)$
$H$ is the height of the binary tree referred by `root`.
# Miscellane
<!--
# Test Cases

[1,null,2,3,4]

[1,2,3,4,5,6,null,null,null,7,8,9,10]

[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]

* Wrong Answer

[3,1,null,null,2,4,null,5]

* Wrong Answer

[1]

* Wrong Answer

[1,null,4,3,null,2,null,5]

-->