LeetCode – Binary Tree Level Order Traversal II (Java)

by

Given a binary tree, return the bottom-up level order traversal of its nodes’ values.

For example, given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as [[15,7], [9,20],[3]]

Java Solution

public List<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
    ArrayList<ArrayList<Integer>> result = new  ArrayList<ArrayList<Integer>>();
 
    if(root == null){
        return result;
    }
 
    LinkedList<TreeNode> current = new LinkedList<TreeNode>();
    LinkedList<TreeNode> next = new LinkedList<TreeNode>();
    current.offer(root);
 
    ArrayList<Integer> numberList = new ArrayList<Integer>();
 
    // need to track when each level starts
    while(!current.isEmpty()){
        TreeNode head = current.poll();
 
        numberList.add(head.val);
 
        if(head.left != null){
            next.offer(head.left);
        }
        if(head.right!= null){
            next.offer(head.right);
        }
 
        if(current.isEmpty()){
            current = next;
            next = new LinkedList<TreeNode>();
            result.add(numberList);
            numberList = new ArrayList<Integer>();
        }
    }
 
    //return Collections.reverse(result);
    ArrayList<ArrayList<Integer>> reversedResult = new  ArrayList<ArrayList<Integer>>();
    for(int i=result.size()-1; i>=0; i--){
        reversedResult.add(result.get(i));
    }
 
    return reversedResult;
}

3 thoughts on “LeetCode – Binary Tree Level Order Traversal II (Java)”

  3. a shortcut is to use the result of leetcode QJ “binary tree level order traversal”, by reversing the result returned by levelOrder~

Leave a Comment