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]
]
和上一题很类似，只不过输出的时候，需要“之”字形输出。
可以在上一层答案的基础上。对偶数层进行反转。
第一次用了两个队列，实现了，但比较慢。
/**
* 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 = new ArrayList<List<Integer>>();
if( root == null)
return list;
Deque<TreeNode> queue = new LinkedList<TreeNode>();
Deque<TreeNode> queue1 = new LinkedList<TreeNode>();
queue.add(root);
int dir = 0;
while( !queue.isEmpty() ){
List ans = new ArrayList<Integer>();
int size = queue.size();
for( int i = 0;i<size;i++){
TreeNode node = null;
if( dir == 0)
node = queue.poll();
else
node = queue.pollLast();
ans.add(node.val);
if( dir == 0){
if( node.left != null)
queue1.add(node.left);
if( node.right != null)
queue1.add(node.right);
}else{
if( node.right != null)
queue1.addFirst(node.right);
if( node.left != null)
queue1.addFirst(node.left);
}
}
list.add(ans);
queue.addAll(queue1);
queue1.clear();
dir = (dir == 1?0:1);
} return list; }
}
但是优点在于不用递归和回溯
使用递归／回溯，速度和代码都会得到优化。
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
helper(res, root, 1);
for (int i = 0; i < res.size(); i++)
if (i % 2 == 1)
Collections.reverse(res.get(i));
return res;
}
private void helper(List<List<Integer>> res, TreeNode node, int level){
if (node == null) return;
if (res.size() < level) res.add(new ArrayList<Integer>());
res.get(level-1).add(node.val);
helper(res, node.left, level+1);
helper(res, node.right, level+1);
}