题目描述
Given a binary tree, return the zigzag level ordertraversal 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] ]
confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1 / \ 2 3 / 4 \ 5The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".
层序遍历,只不过,不同层顺序不同。
参考了别人的代码:用两个stack分别存储不同的两侧,在取stack1里结点的时候,把相应结点的孩子结点放进stack2,同理,取stack2里结点的时候,把相应结点的孩子放进stack1.
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import java.util.*;
public class Solution {
public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
ArrayList<ArrayList<Integer> >matrix=new ArrayList();
if(root==null)
return matrix;
Stack<TreeNode> stack1=new Stack();
Stack<TreeNode> stack2=new Stack();
stack1.push(root);
ArrayList<Integer> array=new ArrayList();
while(!stack1.empty()||!stack2.empty())
{
while(!stack1.empty())//在打印stack1中存放的层时,将下一层存放入stack2
{
TreeNode node=stack1.pop();
if(node.left!=null)
stack2.push(node.left);
if(node.right!=null)
stack2.push(node.right);
array.add(node.val);
if(stack1.empty())//当前array保存完该层结点时,存入matrix,并将array赋给新的ArrayList;
{
matrix.add(array);
array=new ArrayList<Integer>();
}
}
while(!stack2.empty())//在打印stack2中存放的层时,将下一层存放入stack1
{
TreeNode node=stack2.pop();
if(node.right!=null)
stack1.push(node.right);
if(node.left!=null)
stack1.push(node.left);
array.add(node.val);
if(stack2.empty())
{
matrix.add(array);
array=new ArrayList<Integer>();
}
}
}
return matrix;
}
}