leetCode 103.Binary Tree Zigzag Level Order Traversal （二叉树Z字形水平序） 解题思路和方法

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}

,

3
/ \
9  20
/  \
15   7

return its zigzag level order traversal as:

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

思路：与二叉树水平序解题思路差不多，得到水平序结果之后，再对偶数链表反转即可。‘

代码如下：

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
List<List<Integer>> list = new ArrayList<List<Integer>>();
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
dfs(0,root);
//每隔1行交换顺序
for(int i = 1; i < list.size(); i = i+2){
List<Integer> al = list.get(i);
int len = al.size();
//倒序交换
for(int j = 0; j + j < len-1; j++){
int k = al.get(j);
al.set(j, al.get(len-1-j));
al.set(len-1-j, k);
}
}
return list;
}
/**
* 中序遍历，根据深度添加list
* @param dep 树的深度
* @param root 根节点
*/
private void dfs(int dep,TreeNode root){
if(root == null){
return;
}
List<Integer> al;//根据情况得到al值
if(list.size() > dep){
al = list.get(dep);
}else{
al = new ArrayList<Integer>();
list.add(al);
}
dfs(dep+1,root.left);
al.add(root.val);
dfs(dep+1,root.right);
}
}