102.Binary Tree Level Order Traversal

Posted on | In

Tree, Medium

Question

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,null,null,15,7],

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5
  3
 / \
9  20
  /  \
 15   7

return its level order traversal as:

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5
[
  [3],
  [9,20],
  [15,7]
]

Answer

Iterative
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null)
            return res;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()){
            List<Integer> temp = new ArrayList<>();
            
            int size = queue.size();
            for(int i=0;i<size;i++){
                TreeNode node = queue.poll();
                temp.add(node.val);
                if(node.left!=null)
                    queue.offer(node.left);
                if(node.right!=null)
                    queue.offer(node.right);
            }
            res.add(temp);
        }
        return res;
    }
}

Time complexity: O(n), 2ms

Recursive
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null)
            return res;
        recursiveLevel(res, root, 1);
        return res;
    }
    public void recursiveLevel(List<List<Integer>> res, TreeNode node, int depth){
        if(node == null)
            return;
        if(res.size() == depth-1){
            List<Integer> temp1 = new ArrayList<>();
            temp1.add(node.val);
            res.add(temp1);
        }
        else if(res.size() >= depth){
            List<Integer> temp2 = res.get(depth-1);
            temp2.add(node.val);
        }
        recursiveLevel(res, node.left, depth+1);
        recursiveLevel(res, node.right, depth+1);        
    }
}

Time complexity: O(n), 1ms