【LeetCode】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,#,#,15,7}

,

3
/ \
9  20
/  \
15   7

return its zigzag level order traversal as:

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

正负得负，负负得正，

所以每次都是逆序traverse当前的level，要注意的是rightNode和leftNode进入的先后顺序

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
List<TreeNode> current = new ArrayList<>();
if(root == null)
return result;
current.add(root);
List<TreeNode> next = new ArrayList<>();
boolean reverseOrder = true;
while(!current.isEmpty()){
List<Integer> levelResult = new LinkedList<>();
for(int i = current.size() - 1; i >= 0; i--){
TreeNode currentNode = current.get(i);
levelResult.add(0, currentNode.val);
if(reverseOrder){
if(currentNode.right != null) next.add(currentNode.right);
if(currentNode.left != null) next.add(currentNode.left);
}
else{
if(currentNode.left != null) next.add(currentNode.left);
if(currentNode.right != null) next.add(currentNode.right);
}
}
current.clear();
reverseOrder = !reverseOrder;
result.add(levelResult);
List<TreeNode> tmp = current;
current = next;
next = tmp;
}
return result;
}
}