1 题目
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]
]
2 分析
我的思路是修改树的BFS算法，使用两个栈，一个从左到右，一个从右到左存储遍历到的节点, 当节点入栈的同时，将节点中的数值存储到对应的结果数组中。 时间复杂度是
O(n)
,但是空间复杂度是
O(2n)
3 代码
/**
* 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<Integer> tmpRes =
new ArrayList<>();
LinkedList<TreeNode> L2R =
new LinkedList<>();
LinkedList<TreeNode> R2L =
new LinkedList<>();
if(root ==
null){
return result;
}
L2R.push(root);
result.add(
new ArrayList<Integer>(Arrays.asList(root.val)));
while(L2R.size() !=
0 || R2L.size() !=
0){
tmpRes.clear();
while(L2R.size() !=
0){
TreeNode cur = L2R.pop();
if(cur.
right !=
null){
R2L.push(cur.
right);
tmpRes.add(cur.
right.val);
}
if(cur.
left !=
null){
R2L.push(cur.
left);
tmpRes.add(cur.
left.val);
}
}
if(tmpRes.size() !=
0){
result.add(
new ArrayList<Integer>(tmpRes));
tmpRes.clear();
}
while(R2L.size() !=
0){
TreeNode cur = R2L.pop();
if(cur.
left !=
null){
L2R.push(cur.
left);
tmpRes.add(cur.
left.val);
}
if(cur.
right !=
null){
L2R.push(cur.
right);
tmpRes.add(cur.
right.val);
}
}
if(tmpRes.size() !=
0){
result.add(
new ArrayList<Integer>(tmpRes));
tmpRes.clear();
}
}
return result;
}
}
