Leetcode-binary-tree-zigzag-level-order-traversal

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
\
5

The 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;

}
}