Binary Tree Level Order Traversal II 二叉树按层遍历（反向输出）@LeetCode

树的Level Order Traversal加上ArrayList翻转

package Level3;

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;

import Utility.TreeNode;

/**
* Binary Tree Level Order Traversal II
*
*  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,#,#,15,7},
3
/ \
9  20
/  \
15   7
return its bottom-up level order traversal as:
[
[15,7]
[9,20],
[3],
]
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}".
*
*/
public class S107 {

public static void main(String[] args) {

}

public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
ArrayList<ArrayList<Integer>> ret = new ArrayList<ArrayList<Integer>>();

if(root == null){
return ret;
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);

ArrayList<ArrayList<Integer>> alal = new ArrayList<ArrayList<Integer>>();
ArrayList<Integer> al = new ArrayList<Integer>();

// 记录两个当前level和下一个level的node数目
int currentLevel = 1;
int nextLevel = 0;

while( !queue.isEmpty() ){
TreeNode cur = queue.remove();
currentLevel--;
al.add(cur.val);

if(cur.left != null){
queue.add(cur.left);
nextLevel++;
}
if(cur.right != null){
queue.add(cur.right);
nextLevel++;
}

if(currentLevel == 0){
alal.add(al);
al = new ArrayList<Integer>();
currentLevel = nextLevel;
nextLevel = 0;
}
}

// 翻转ArrayList
for(int i=alal.size()-1; i>=0; i--){
ret.add(alal.get(i));
}

return ret;
}

}

/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
ArrayList<ArrayList<Integer>> ret = new ArrayList<ArrayList<Integer>>();
if(root == null){
return ret;
}
LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
int curLevCnt = 1;
int nextLevCnt = 0;
ArrayList<Integer> al = new ArrayList<Integer>();
while(!queue.isEmpty()){
TreeNode cur = queue.remove();
curLevCnt--;
al.add(cur.val);
if(cur.left != null){
queue.add(cur.left);
nextLevCnt++;
}
if(cur.right != null){
queue.add(cur.right);
nextLevCnt++;
}

if(curLevCnt == 0){
curLevCnt = nextLevCnt;
nextLevCnt = 0;
ret.add(al);
al = new ArrayList<Integer>();
}
}
ArrayList<ArrayList<Integer>> ret2 = new ArrayList<ArrayList<Integer>>();
for(int i=ret.size()-1; i>=0; i--){
ret2.add(ret.get(i));
}

return ret2;
}
}