[LeetCode]Binary Tree Zigzag Level Order Traversal

 1 public class Solution {
 2     public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
 3         List<List<Integer>> result = new ArrayList<List<Integer>>();
 4         if (root == null) {
 5             return result;
 6         }
 7         Stack<TreeNode> stack = new Stack<TreeNode>();
 8         stack.push(root);
 9         boolean flg = true;
10         while (!stack.isEmpty()) {
11             List<Integer> list = new ArrayList<Integer>();
12             Queue<TreeNode> queue = new LinkedList<TreeNode>();
13             while (!stack.isEmpty()) {
14                 TreeNode node = stack.pop();
15                 list.add(node.val);
16                 if (flg) {
17                     if (node.left != null) {
18                         queue.offer(node.left);
19                     }
20                     if (node.right != null) {
21                         queue.offer(node.right);
22                     }
23                 } else {
24                     if (node.right != null) {
25                         queue.offer(node.right);
26                     }
27                     if (node.left != null) {
28                         queue.offer(node.left);
29                     }
30                 }
31             }
32             while (!queue.isEmpty()) {
33                 stack.push(queue.poll());
34             }
35             flg = !flg;
36             result.add(list);
37         }
38         return result;
39     }
40 }

 下面的方法代码复杂度低一些

 1 public class Solution {
 2     public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
 3         List<List<Integer>> result = new ArrayList<List<Integer>>();
 4         if (root == null) {
 5             return result;
 6         }
 7         boolean flg = false;
 8         Queue<TreeNode> queue = new LinkedList<TreeNode>();
 9         queue.offer(root);
10         while (!queue.isEmpty()) {
11             List<Integer> list = new LinkedList<Integer>();
12             int size = queue.size();
13             for (int i = 0; i < size; i++) {
14                 TreeNode node = queue.poll();
15                 if (flg) {
16                     list.add(0, node.val);
17                 } else {
18                     list.add(node.val);
19                 }
20                 if (node.left != null) {
21                     queue.offer(node.left);
22                 }
23                 if (node.right != null) {
24                     queue.offer(node.right);
25                 }
26             }
27             result.add(list);
28             flg = !flg;
29         }
30         return result;
31     }
32 }

 

转载于:https://www.cnblogs.com/vision-love-programming/p/5018362.html