LeetCode 545: Boundary of Binary Tree Solution

Master LeetCode problem 545 (Boundary of Binary Tree), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

545. Boundary of Binary Tree

Medium

Tree Depth-First Search Binary Tree

Problem Explanation

Explanation:

The problem "Boundary of Binary Tree" asks us to return the boundary of a binary tree in anti-clockwise order starting from the root. The boundary of a binary tree consists of four parts: the left boundary, the leaves, the right boundary, and the root.

To solve this problem, we can follow these steps:

Traverse the left boundary in a top-down manner until we reach the leftmost leaf node.
Traverse the leaf nodes from left to right.
Traverse the right boundary in a bottom-up manner starting from the rightmost leaf node.
Construct the boundary by combining these three parts.

Time Complexity: O(n) where n is the number of nodes in the binary tree. Space Complexity: O(n) for the recursive function calls and the output boundary list.

Solution Code

class Solution {
    public List<Integer> boundaryOfBinaryTree(TreeNode root) {
        List<Integer> boundary = new ArrayList<>();
        if (root == null) {
            return boundary;
        }
        
        boundary.add(root.val);
        if (root.left != null) {
            getLeftBoundary(root.left, boundary);
        }
        getLeaves(root, boundary);
        if (root.right != null) {
            getRightBoundary(root.right, boundary);
        }
        
        return boundary;
    }
    
    private void getLeftBoundary(TreeNode node, List<Integer> boundary) {
        if (node == null || (node.left == null && node.right == null)) {
            return;
        }
        
        boundary.add(node.val);
        if (node.left == null) {
            getLeftBoundary(node.right, boundary);
        } else {
            getLeftBoundary(node.left, boundary);
        }
    }
    
    private void getLeaves(TreeNode node, List<Integer> boundary) {
        if (node == null) {
            return;
        }
        
        if (node.left == null && node.right == null) {
            boundary.add(node.val);
        }
        
        getLeaves(node.left, boundary);
        getLeaves(node.right, boundary);
    }
    
    private void getRightBoundary(TreeNode node, List<Integer> boundary) {
        if (node == null || (node.left == null && node.right == null)) {
            return;
        }
        
        if (node.right == null) {
            getRightBoundary(node.left, boundary);
        } else {
            getRightBoundary(node.right, boundary);
        }
        boundary.add(node.val);
    }
}

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