Binary Tree Level Order Traversal

Fair Binary Tree Level Order TraversalMy Submissions

30%

Accepted

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

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]
]

1. Use two queues.

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
 
 
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Level order a list of lists of integer
     */
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> rst = new ArrayList<ArrayList<Integer>>();
        if(root == null) {
            return rst;
        }
        ArrayList<Integer> level = new ArrayList<Integer>();
        Queue<TreeNode> cur = new LinkedList<TreeNode>();
        Queue<TreeNode> next = new LinkedList<TreeNode>();
        cur.add(root);
        while (!cur.isEmpty()) {
            TreeNode temp = cur.poll();
            level.add(temp.val);
            if (temp.left != null) {
                next.add(temp.left);
            } 
            if (temp.right != null) {
                next.add(temp.right);
            }
            if(cur.isEmpty()) {
                rst.add(new ArrayList(level));
                level = new ArrayList<Integer>();
                if (!next.isEmpty()) {
                    cur = next;
                    next = new LinkedList<TreeNode>();
                }
            }
        }
        return rst;
    }
}

2. Use one queue, use a counter to record the number of nodes in each level

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
 
 
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Level order a list of lists of integer
     */
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> rst = new ArrayList<ArrayList<Integer>>();
        if(root == null) {
            return rst;
        }
        ArrayList<Integer> level = new ArrayList<Integer>();
        Queue<TreeNode> cur = new LinkedList<TreeNode>();
        cur.add(root);
        while (!cur.isEmpty()) {
            int size = cur.size();
            for(int i = 0; i < size; i++) {
                TreeNode temp = cur.poll();
                level.add(temp.val);
                if (temp.left != null) {
                    cur.add(temp.left);
                } 
                if (temp.right != null) {
                    cur.add(temp.right);
                }
            }
            rst.add(new ArrayList(level));
            level = new ArrayList<Integer>();
        }
        return rst;
    }
}