103. Binary Tree Zigzag Level Order Traversal

class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> res=new ArrayList<List<Integer>>();
        if(root==null)
            return res;
        Queue<TreeNode> queue=new LinkedList<TreeNode>();
        List<Integer> list=new ArrayList<Integer>();
        queue.add(root);
        queue.add(null);
        boolean zig=true;
        while(!queue.isEmpty())
        {
            TreeNode node=queue.poll();
            if(node==null)
            {
                res.add(new ArrayList<Integer>(list));
                if(!queue.isEmpty())
                    queue.add(node);
                list.clear();
                zig=!zig;
            }
            else
            {
                if(zig==true)
                    list.add(node.val);
                else
                    list.add(0, node.val);
                if(node.left!=null)
                    queue.add(node.left);
                if(node.right!=null)
                    queue.add(node.right);
            }
        }
        return res;
    }
}

  

原文地址:https://www.cnblogs.com/asuran/p/7610312.html