Zigzag Level Order Traversal
出处
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]
]
Solution
- Queue + reverse.
- Two stacks.
Complexity
时间复杂度 O(n),空间复杂度 O(n)
Code
public class Solution {
public List<List<Integer>> zigzagLevelOrder_1(TreeNode root) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if (root == null) return res;
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
q.offer(null);
List<Integer> level = new ArrayList<Integer>();
int depth = 0;
while(true) {
TreeNode node = q.poll();
if (node != null) {
level.add(node.val);
if(node.left!=null) q.offer(node.left);
if(node.right!=null) q.offer(node.right);
} else {
if (depth % 2 == 1) Collections.reverse(level);
res.add(level);
depth++;
level = new ArrayList<Integer>();
if(q.isEmpty()==true) break;
q.offer(null);
}
}
return res;
}
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if (root == null) return res;
Stack<TreeNode> cur = new Stack<TreeNode>();
Stack<TreeNode> last = new Stack<TreeNode>();
boolean left2right = true;
last.push(root);
List<Integer> level = new ArrayList<Integer>();
while (last.empty() == false) {
TreeNode node = last.pop();
if (node != null) {
level.add(node.val);
if (left2right) {
if(node.left!=null) cur.push(node.left);
if(node.right!=null) cur.push(node.right);
} else {
if(node.right!=null) cur.push(node.right);
if(node.left!=null) cur.push(node.left);
}
}
if (last.empty() == true) {
if (level.size() != 0)
res.add(level);
level = new ArrayList<Integer>();
Stack<TreeNode> temp = last;
last = cur;
cur = temp;
left2right = !left2right;
}
}
return res;
}
}