Binary Tree Zigzag Level Order Traversal
3
/ \
9 20
/ \
15 7[
[3],
[20,9],
[15,7]
]Last updated
3
/ \
9 20
/ \
15 7[
[3],
[20,9],
[15,7]
]Last updated
/**
* 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;
}
}