Binary Tree Level Order Traversal

Problem Statement

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

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

3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

Approach

• Starting from the root, store nodes at each level in a TreeMap<Level,List<Nodes>>
• A TreeMap is used to maintain the natural ordering from level 0 to n
• Iterate over the Map to return all the lists of nodes for each level.

Run Time Complexity

The run time for this traversal is Theta(n)

Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    Map<Integer,List<Integer>> map = new TreeMap<>();
    List<List<Integer>> result = new ArrayList<>();
    
    public List<List<Integer>> levelOrder(TreeNode root) {
        if(root == null){
            return result;
        }
        updateMap(root,-1);
        for(Map.Entry<Integer,List<Integer>> entry: map.entrySet()){
            result.add(entry.getValue());
        }
        return result;
    }
    
    private void updateMap(TreeNode root,int level){
        if(root == null){
            return;
        }
        level++;
        if(map.containsKey(level)){
            map.get(level).add(root.val);
            
        }else{
            List<Integer> list = new ArrayList<>();
            list.add(root.val);
            map.put(level,list);    
        }
        updateMap(root.left,level);
        updateMap(root.right,level);
        return;
    }
    
}