[LeetCode刷题]Binary Tree Level Order Traversal

看到LinkedIn有考这道题目，就做了一下。 没什么特别的，在queue里面加了一个null来表示是在当前level的尽头

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> ret = new LinkedList<List<Integer>>();
        if(root == null) {
            return ret;
        }
        //The Queue for bfs
        LinkedList<TreeNode> q = new LinkedList<TreeNode>();
        q.add(root);
        q.add(null);//null means we are at the end of a level

        List<Integer> l = new LinkedList<Integer>();
        //这里开始traverse
        while(q.size() > 0) {
            TreeNode cur = q.poll();
            l.add(cur.val);
            if(cur.left != null) {
                q.add(cur.left);
            }
            if(cur.right != null) {
                q.add(cur.right);
            }
            if(q.peek() == null) {//meaning we have reached the end of a level
                ret.add(l);
                l = new LinkedList<Integer>();
                //表示是level的最后
                q.add(null);
                while(q.peek() == null && q.size() > 0) {
                    q.poll();
                }
            }
        }
        return ret;
    }
}

看到有人说还要考dfs的traverse， 尿了，赶紧乖乖奉上dfs的解法

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        //这里不用考虑edge case，直接在helper function里面考虑
        List<List<Integer>> ret = new LinkedList<List<Integer>>();
        _levelOrder(root, ret, 0);
        return ret;
    }
    public void _levelOrder(TreeNode cur, List<List<Integer>> l, int depth) {
        if(cur == null) {
            return;
        }
        if(l.size() >= (depth+1)) {
            List<Integer> temp = l.get(depth);
            temp.add(cur.val);
        } else {
            List<Integer> temp = new LinkedList<Integer>();
            temp.add(cur.val);
            l.add(temp);
        }
        //从左到右，可以保证按顺序。用depth记录当前的深度
        _levelOrder(cur.left, l, depth+1);
        _levelOrder(cur.right, l, depth+1);
    }
}

两个queue的解法，这里是print

public void printTree(TreeNode tmpRoot) {

        Queue<TreeNode> currentLevel = new LinkedList<TreeNode>();
        Queue<TreeNode> nextLevel = new LinkedList<TreeNode>();

        currentLevel.add(tmpRoot);

        while (!currentLevel.isEmpty()) {
            Iterator<TreeNode> iter = currentLevel.iterator();
            while (iter.hasNext()) {
                TreeNode currentNode = iter.next();
                if (currentNode.left != null) {
                    nextLevel.add(currentNode.left);
                }
                if (currentNode.right != null) {
                    nextLevel.add(currentNode.right);
                }
                System.out.print(currentNode.value + " ");
            }
            System.out.println();
            currentLevel = nextLevel;
            nextLevel = new LinkedList<TreeNode>();

        }

    }