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] ]
要求之字形輸出節點值 即每隔行的輸出順序相反 很明顯這與棧的定義相似 用棧來存儲則每回輸出的內容剛好與此前相反 若用一個棧則不能存儲新的節點 所以 用兩個棧 注意的是 從右向左輸出時 添加節點的順序也要改變爲右向左 代碼如下：
public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> res=new ArrayList<List<Integer>>();
if(root==null) return res;
int count=0;
int level=1;
int num=1;
Stack<TreeNode> sta2=new Stack<TreeNode>();
sta2.push(root);
Stack<TreeNode> sta=new Stack<TreeNode>();
while(sta2.isEmpty()!=true||sta.isEmpty()!=true){
List<Integer> tmp=new ArrayList<Integer>();
num=level;
level=0;
if(count%2==0){
for(int i=0;i<num;i++){
tmp.add(sta2.peek().val);
if(sta2.peek().left!=null){
sta.add(sta2.peek().left);
level++;
}
if(sta2.peek().right!=null){
sta.add(sta2.peek().right);
level++;
}
sta2.pop();
}
}
if(count%2==1){
for(int i=0;i<num;i++){
tmp.add(sta.peek().val);
if(sta.peek().right!=null){
sta2.add(sta.peek().right);
level++;
}
if(sta.peek().left!=null){
sta2.add(sta.peek().left);
level++;
}
sta.pop();
}
}
res.add(tmp);
count++;
}
return res;
}
}