Binary Tree Zigzag Level Order Traversal

Binary Tree Zigzag Level Order Traversal

问题：

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

思路：

　　队列层次访问

我的代码：

public class Solution {

    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {

        List<List<Integer>> rst = new ArrayList<List<Integer>>();

        if(root == null)    return rst;

        Queue<TreeNode> queue = new LinkedList<TreeNode>();

        queue.offer(root);

        int count = 1;

        while(!queue.isEmpty())

        {

            int size = queue.size();

            List<Integer> list = new ArrayList<Integer>();

            for(int i = 0; i < size; i++)

            {

                TreeNode node = queue.poll();

                if(node.left != null)   queue.offer(node.left);

                if(node.right != null)  queue.offer(node.right);

                list.add(node.val);

            }

            if(count%2 == 1)

            {

                rst.add(list);

            }

            else

            {

                Collections.reverse(list);

                rst.add(list);

            }

            count++;

        }

        return rst;

    }

}

View Code

他人代码：

public class Solution {

    public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {

        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();



        if (root == null) {

            return result;

        }



        Stack<TreeNode> currLevel = new Stack<TreeNode>();

        Stack<TreeNode> nextLevel = new Stack<TreeNode>();

        Stack<TreeNode> tmp;

        

        currLevel.push(root);

        boolean normalOrder = true;



        while (!currLevel.isEmpty()) {

            ArrayList<Integer> currLevelResult = new ArrayList<Integer>();



            while (!currLevel.isEmpty()) {

                TreeNode node = currLevel.pop();

                currLevelResult.add(node.val);



                if (normalOrder) {

                    if (node.left != null) {

                        nextLevel.push(node.left);

                    }

                    if (node.right != null) {

                        nextLevel.push(node.right);

                    }

                } else {

                    if (node.right != null) {

                        nextLevel.push(node.right);

                    }

                    if (node.left != null) {

                        nextLevel.push(node.left);

                    }

                }

            }



            result.add(currLevelResult);

            tmp = currLevel;

            currLevel = nextLevel;

            nextLevel = tmp;

            normalOrder = !normalOrder;

        }



        return result;



    }

}

View Code

学习之处：

• 虽然使用队列也能实现该问题，但是仔细想想，如果这一层的最右节点是这层的最后一个元素，那么它的right节点是下一层的最后一个访问的，如果这一层的最左节点是该层访问的最后一个元素，那么该元素的Left将是下一层访问的节点，故由此分析，使用Stack比queue要更加合理，而且如此一来，也能节约掉Queue方法下面的链表翻转的时间。