给定一个二叉树,返回其节点值的锯齿形层次遍历。(即先从左往右,再从右往左进行下一层遍历,以此类推,层与层之间交替进行)。
例如:
给定二叉树 [3,9,20,null,null,15,7]
,
3 / \ 9 20 / \ 15 7
返回锯齿形层次遍历如下:
[ [3], [20,9], [15,7] ]
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] ]
个人思路:
以二叉树的层次遍历的思路为基础,定义direction作为存储方向(true代表向右存储,false代表向左存储),以满足锯齿形存储的要求,仍然按层次遍历取得子数组,每次内循环获得一层的元素,并在每一轮结束后根据direction的布尔值进行顺序存储或反向存储,构成新的子数组加入总数组,最终得到锯齿形层次遍历数组。
代码(Java):
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
Queue<TreeNode> queue=new LinkedList<TreeNode>();
List<List<Integer>> visited=new LinkedList<List<Integer>>();
int row=0,col=0;
Boolean direction=true;//true代表向右,false代表向左
if(root!=null){queue.offer(root);}
else{return visited;}
while(!queue.isEmpty()){
int level=queue.size();
List<Integer> sub_visited=new LinkedList<Integer>();
Stack<Integer> sub_stack=new Stack<Integer>();
List<Integer> sub_queue=new LinkedList<Integer>();
for(int i=0;i<level;i++){
if(queue.peek().left!=null){queue.offer(queue.peek().left);}
if(queue.peek().right!=null){queue.offer(queue.peek().right);}
int value=queue.poll().val;
sub_stack.push(value);
sub_queue.add(value);
}
if(direction==true){
sub_visited=sub_queue;
}
else{
while(!sub_stack.isEmpty()){
sub_visited.add(sub_stack.pop());
}
}
direction=!direction;
visited.add(sub_visited);
}
return visited;
}
}