LeetCode力扣之103. Binary Tree Zigzag Level Order Traversal

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).For example:
Given binary tree 

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

,
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]

package leetCode;

import java.util.ArrayList;
import java.util.Deque;
import java.util.LinkedList;
import java.util.List;

/**
* Created by lxw, liwei4939@126.com on 2018/3/21.
*/
public class L103_BinaryTreeZigZagOrderTravel {

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

Deque<TreeNode> dq = new LinkedList<>();
List<List<Integer>> res = new ArrayList<>();
if (root == null){
return res;
}

boolean lr = true;
TreeNode last = root;
TreeNode nlast = null;
dq.addFirst(root);
List<Integer> lay = new ArrayList<>();

while (!dq.isEmpty()){

if (lr){
root = dq.pollFirst();
lay.add(root.val);

if (root.left != null){
nlast = nlast == null? root.left : nlast;
dq.offerLast(root.left);
}

if (root.right != null){
nlast = nlast == null ? root.right : nlast;
dq.offerLast(root.right);
}
} else {
root = dq.pollLast();
lay.add(root.val);

if (root.right != null){
nlast = nlast == null ? root.right : nlast;
dq.offerFirst(root.right);
}

if (root.left != null){
nlast = nlast == null ? root.left : nlast;
dq.offerFirst(root.left);
}
}

if (root == last){
lr = !lr;
last = nlast;
nlast = null;
res.add(lay);
lay = new ArrayList<>();
}
}

return res;
}

public static void main(String[] args){
L103_BinaryTreeZigZagOrderTravel tmp = new L103_BinaryTreeZigZagOrderTravel();
TreeNode root = new TreeNode(3);
root.left = new TreeNode(9);
root.right = new TreeNode(20);

root.right.left = new TreeNode(15);
root.right.right = new
ae17
TreeNode(7);

List<List<Integer>> res = tmp.zigzagLevelOrder(root);
System.out.println(res);
}
}