Binary Tree Zigzag Level Order Traversal - LeetCode

最新推荐文章于 2020-07-25 07:42:38 发布

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最新推荐文章于 2020-07-25 07:42:38 发布

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探讨二叉树的zigzag level order遍历方法，包括遍历算法实现及特殊情况处理。

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Binary Tree Zigzag Level Order Traversal

Total Accepted: 5345 Total Submissions: 20535 My Submissions

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},

return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]

这道题目我认为我正确了，但是没有AC,大体例子都过了，但是出现错误如下，不知道为什么。

Input:	{1}
Output:	[1]
Expected:	[[1]]

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
     static int index = 0;
	 static int currentlevel = 1;
	 static int nextlevel = 0;
	 static int count = 0;
	 static ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
	 static ArrayList<Integer> r = new ArrayList<Integer>();
	 static Queue<TreeNode> s = new LinkedList<TreeNode>();
	 
	 public void traverse(TreeNode root){
			 
		 while(!s.isEmpty()){
			 TreeNode current = s.remove();
			 r.add(current.val);
			 count ++;
			 if(current.left != null){
			 s.add(current.left);
			 nextlevel ++;
			 }
			 if(current.right != null){
				 s.add(current.right);
				 nextlevel ++;
			 }
			if(currentlevel == count)
			{
				ArrayList<Integer> rr = new ArrayList<Integer>();
				rr = r;
				if(index % 2 == 1){
					Collections.reverse(rr);
					result.add(rr);
					r = new ArrayList<Integer>();
				}
				else{
					result.add(rr);
					r = new ArrayList<Integer>();
				}
				System.out.println(rr);
				count = 0;
				currentlevel = nextlevel;
				index = index + 1;
				nextlevel = 0;
			}
			 
		 }
	 }
    public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
        if(root == null)
        return result;
         s.add(root);  
		 traverse(root);
	       
			return result;
    }
}