leetcode103JAVA_[LeetCode] 103. Binary Tree Zigzag Level Order Traversal Java

题目：

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],

/ \

9 20

/ \

15 7

return its zigzag level order traversal as:

[

[3],

[20,9],

[15,7]

]

题意及分析：给出一个横向遍历的二叉树，要求给出按照Z字形横向遍历的每一层的点。这道题我们可以用stack来求解，简单来讲就是把每一层添加到一个stack里面，然后遍历输出结果的同时将遍历点的子节点添加到一个新的stack里面(先添加左子节点还是右子节点根据上一层添加的顺序)，遍历完之后将新的stack赋值给stack进行下一层的遍历。代码如下：

代码：

/**

* 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> zigzagLevelOrder(TreeNode root) {

List> res=new ArrayList<>();

TreeNode node=root;

Stack stack =new Stack<>();

if(node!=null)

stack.push(node);

int leftOrRight=0;//0为先遍历左子树，1为右子树

while(!stack.isEmpty()){

Stack nextStack =new Stack<>();

List list=new ArrayList<>();

while(!stack.isEmpty()){

TreeNode temp=stack.pop();

list.add(temp.val);

if(leftOrRight==0){

if(temp.left!=null)

nextStack.push(temp.left);

if(temp.right!=null)

nextStack.push(temp.right);

}else{

if(temp.right!=null)

nextStack.push(temp.right);

if(temp.left!=null)

nextStack.push(temp.left);

}

if(leftOrRight==0)

leftOrRight=1;

else {

leftOrRight=0;

}

res.add(new ArrayList<>(list));

//System.out.println(list);

list.clear();

stack=(Stack) nextStack.clone ();;

nextStack.clear();

}

return res;

}

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