【leetcode】103. Binary Tree Zigzag Level Order Traversal【java】使用队列，简单容易理解的方法

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]
]

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use a queue to implement BFS. Each time when I poll a node, I add this node value to level. I use a variable
zigzag to indicate whether add from left to right or right to left. If zigzag == false, it is from left to right; if zigzag == true, it is from right to left.
And from right to left just need to use ArrayList.add(0, value) method

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
//非递归：使用队列
public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> res = new LinkedList<List<Integer>>();
if (root == null){
return res;
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
boolean isZigzag = false;
while (!queue.isEmpty()){
List<Integer> subList = new LinkedList<Integer>();
int count = queue.size();
for (int i = 0; i < count; i++){
TreeNode node = queue.poll();
if (isZigzag){
subList.add(0,node.val);
} else {
subList.add(node.val);
}
if (node.left != null){
queue.add(node.left);
}
if (node.right != null){
queue.add(node.right);
}
}
res.add(subList);
isZigzag = !isZigzag;
}
return res;
}
}