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,null,null,15,7]

,

3
/ \
9  20
/  \
15   7

return its bottom-up level order traversal as:

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

对树的逆层次遍历。我们不可能从底部往上走，所以我们至顶向下走，不断的在链表前端插入新的一层节点。

public class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
int cnt = 0;
int f = 0;
int t = 1;
int index;
List<List<Integer>> lsts = new ArrayList<List<Integer>>();
List<Integer> lst = new ArrayList<Integer>();
List<TreeNode> nodeList = new ArrayList<TreeNode>();
nodeList.add(root);
if(root == null){
return lsts;
}
index = 0;
while(t != 0){
if(f == t){
t = cnt;
f = 0;
cnt = 0;
lsts.add(0,lst);
lst = new ArrayList<Integer>();
}
else{
TreeNode node = nodeList.get(index ++);
if(node != null){
f ++;
lst.add(node.val);
}
if(node.left != null){
cnt ++;
nodeList.add(node.left);
}
if(node.right != null){
cnt ++;
nodeList.add(node.right);
}
}
}
return lsts;
}
}
然而只占胜了23.52%，于是我们选择还是按顺序放入链表，在进行反转。

public class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
int cnt = 0;
int f = 0;
int t = 1;
int index;
List<List<Integer>> lsts = new ArrayList<List<Integer>>();
List<Integer> lst = new ArrayList<Integer>();
List<TreeNode> nodeList = new ArrayList<TreeNode>();
nodeList.add(root);
if(root == null){
return lsts;
}
index = 0;
while(t != 0){
if(f == t){
t = cnt;
f = 0;
cnt = 0;
lsts.add(lst);
lst = new ArrayList<Integer>();
}
else{
TreeNode node = nodeList.get(index ++);
if(node != null){
f ++;
lst.add(node.val);
}
if(node.left != null){
cnt ++;
nodeList.add(node.left);
}
if(node.right != null){
cnt ++;
nodeList.add(node.right);
}
}
}
int from = 0,to = lsts.size() - 1;
while(from < to){
List<Integer> tempF = lsts.get(from);
List<Integer> tempE = lsts.get(to);
lsts.set(from, tempE);
lsts.set(to, tempF);
from ++;
to --;
}
return lsts;
}
}

此刻战胜了69.49% 。当然了，自然是可以利用递归解决的，递归也可以解决的。
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> wrapList = new LinkedList<List<Integer>>();
levelMaker(wrapList, root, 0);
return wrapList;
}

public void levelMaker(List<List<Integer>> list, TreeNode root, int level) {
if(root == null) return;
if(level >= list.size()) {
list.add(0, new LinkedList<Integer>());
}
levelMaker(list, root.left, level+1);
levelMaker(list, root.right, level+1);
list.get(list.size()-level-1).add(root.val);
}

这个方案战胜了69.49%