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  • [leedcode 103] Binary Tree Zigzag Level Order Traversal

    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]
    ]
    /**
     * 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) {
            //和正常的层序遍历的唯一不同是,增加一个标志flag,代表从左还是从右开始遍历,使用了Collections的静态方法reverse反转链表
            List<List<Integer>> res=new ArrayList<List<Integer>>();
            List<Integer> seq=new ArrayList<Integer>();
            if(root==null) return res;
            LinkedList<TreeNode> queue=new LinkedList<TreeNode>();
            queue.add(root);
            int cutcont=1;
            int nextcont=0;
            boolean flag=false;
            while(!queue.isEmpty()){
                  TreeNode node=queue.remove();
                 seq.add(node.val);
                 cutcont--;
                if(node.left!=null){
                    queue.add(node.left);
                    nextcont++;
                }
                if(node.right!=null){
                    queue.add(node.right);
                    nextcont++;
                }
             /*   while(cutcont>0){
                    TreeNode node=queue.remove();
                    seq.add(node.val);
                     cutcont--;
                    if(node.left!=null){
                    queue.add(node.left);
                    nextcont++;
                    }
                    if(node.right!=null){
                    queue.add(node.right);
                    nextcont++;
                }
                }*/
                
                if(cutcont==0){
                    if(flag){
                        Collections.reverse(seq);
                        
                    }
                    flag=!flag;
                    res.add(seq);
                    cutcont=nextcont;
                    nextcont=0;
                    seq=new ArrayList<Integer>();
                    
                }
            }
            return res;
            
            
        }
        
    }
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  • 原文地址:https://www.cnblogs.com/qiaomu/p/4662811.html
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