LinkedIn Binary Tree Zigzag Level Order Traversal 每次往res里放list的时候,要check一下list是不是空,因为很有可能树遍历完以后,有一个stack是空的,那么对应的list就是空的,空的list不能加到最后的接过去
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]
,
return its zigzag level order traversal as:
Copy [
[3],
[20,9],
[15,7]
]
用两个栈维护正序和反序输出,s1往s2push 的时候,左先进,右后进,这样在遍历s2时,就先遍历后进去的,输出的顺序就是倒着的层序遍历,s2往s1 push的时候,右先进,左后进,这样遍历的时候就是从左到右的正常循序,
2个坑:
用同一个list的话,往res里传的时候应该deep copy
往res里传的时候应该保证list不为空,因为两个stack的循环都在一个大循环里,有时候期中一个stack为空
时间复杂度 o(n) , space o(n)
Copy /**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if(root == null)
return res;
Stack<TreeNode> s1 = new Stack<>();
Stack<TreeNode> s2 = new Stack<>();
s1.push(root);
while(!s1.isEmpty() || !s2.isEmpty()){
List<Integer> list = new ArrayList<>();
while(!s1.isEmpty()){
TreeNode node = s1.pop();
list.add(node.val);
//先进左,后进右,右先出,左后出
if(node.left != null) s2.push(node.left);
if(node.right != null) s2.push(node.right);
}
if(!list.isEmpty())
res.add(new ArrayList<>(list));
list.clear();
while(!s2.isEmpty()){
TreeNode node = s2.pop();
list.add(node.val);
//右先进,左后进,左先出,右后出
if(node.right != null) s1.push(node.right);
if(node.left != null) s1.push(node.left);
}
if(!list.isEmpty())
res.add(new ArrayList<>(list));
list.clear();
}
return res;
}
}