545. Boundary of Binary Tree

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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