Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree [3,9,20,null,null,15,7],
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
解題思路:
二叉樹的層次遍歷
用兩個List分別暫時存儲奇數層的節點和偶數層的節點,奇數層List中開始存儲的是root節點,然後把奇數層節點的子節點依次全部放入存儲偶數層的節點List中,並在存儲的過程中把子節點存儲過的節點刪掉並把奇數層幾點的值存儲在一個List中,然後進行偶數層遍歷,循環至兩個List都爲空表示遍歷完成。
具體代碼實現如下:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> result = new ArrayList();
if(root == null){
return result;
}
List<TreeNode> s_even = new ArrayList();
s_even.add(root);
List<TreeNode> s_odd = new ArrayList();
int i = 1;//記錄進行到是奇數層還是偶數層
while(true){
List<Integer> list = new ArrayList();//存儲一層中的所有數據
if(i%2 == 1){
while(s_even.isEmpty() == false){
TreeNode node = s_even.get(0);
s_even.remove(0);
if(node.left != null){
s_odd.add(node.left);
}
if(node.right != null){
s_odd.add(node.right);
}
list.add(node.val);//把這個節點數據加入到list中
}
}
if(i % 2 == 0){
while(s_odd.isEmpty() == false){
TreeNode node = s_odd.get(0);
s_odd.remove(node);
if(node.left != null){
s_even.add(node.left);
}
if(node.right != null){
s_even.add(node.right);
}
list.add(node.val);
}
}
result.add(list);//把一層的結果加入到最終結果中
if(s_even.isEmpty()&s_odd.isEmpty()){//結束條件
break;
}
i++;
}
List<List<Integer>> result_temp = new ArrayList();
int re_len = result.size();
//因爲題目需要的最終結果是從最底層開始的,而我們是從頂開始的 所以進行逆序
for(int j = 0; j < re_len; j++){
result_temp.add(result.get(re_len-1-j));
}
return result_temp;
}
}