103. Binary Tree Zigzag Level Order Traversal

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难点：Z字形，奇数行从左向右输出，偶数行从右向左输出（但是注意这个偶数行的下一行又要以从左向右输出）。
解决思路：用栈和队列来解决顺序和逆序的问题。
1.用phase来表示奇数行还是偶数行；
2.对于奇数行用队列nodeQueue来记录所有点，对于偶数行，用nodeStack来记录所有点（从左到右push进去，则出来时从右到左），偶数行的输出用nodeStack弹出的值来完成；与nodeStack对应还有一个nodeQueueForSeq，用来记录偶数行的下一行的节点。

代码如下：

public class Solution {    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {        List<List<Integer>> nodeLevels = new ArrayList<List<Integer>>();        if (root == null) return nodeLevels;        Stack<TreeNode> nodeStack = new Stack<TreeNode>();  // nodes output from right to left        Queue<TreeNode> nodeQueueForSeq = new LinkedList<TreeNode>();  //corresponding queue for nodeStack, because the next generation of this level should be from left to right        Queue<TreeNode> nodeQueue = new LinkedList<TreeNode>();  // nodes output from left to right        boolean phase = true;   // true: from left to right; false: from right to left        TreeNode node = null;        TreeNode nodeForSeq = null;        nodeQueue.offer(root);        while(!nodeStack.empty() || nodeQueue.size() != 0){            List<Integer> nodeLevel = new ArrayList<Integer>();            if (phase == true){                 while(nodeQueue.size() != 0){                    node = nodeQueue.poll();                    nodeLevel.add(node.val);                    if (node.left != null) {                        nodeStack.push(node.left);                        nodeQueueForSeq.offer(node.left);                    }                    if (node.right != null) {                        nodeStack.push(node.right);                        nodeQueueForSeq.offer(node.right);                    }                }            }else{                while(!nodeStack.empty()){                    node = nodeStack.pop();                    nodeLevel.add(node.val);                    nodeForSeq = nodeQueueForSeq.poll();                    if (nodeForSeq.left != null) nodeQueue.offer(nodeForSeq.left);                    if (nodeForSeq.right != null) nodeQueue.offer(nodeForSeq.right);                    //nodeLevel.add(nodeForValue.val);                }            }            nodeLevels.add(nodeLevel);            phase = !phase;        }        return nodeLevels;    }}

解法二：找到一个更好更简单的方法(https://discuss.leetcode.com/topic/59174/1ms-short-java-solution-using-depth)

public class Solution {    // leftDirection = true means adding elements from left to right    void dfs(TreeNode root, List<List<Integer>> result, int depth, boolean leftDirection) {        if(root==null) return;        if(result.size()==depth){            result.add(new LinkedList<Integer>());        }        List<Integer> cur = result.get(depth);        if(!leftDirection) cur.add(0, root.val);        else cur.add(root.val);        dfs(root.left, result, depth+1, !leftDirection);        dfs(root.right, result, depth+1, !leftDirection);    }    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {        List<List<Integer>> result = new ArrayList<List<Integer>>();        dfs(root, result, 0, true);        return result;    }}

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