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]
]

解题思路：

首先想到的是二叉树的层序遍历，此题有两个关注点，首先是如何用两个计数变量来确定二叉树每一层上子结点的个数，其次是Arraylist.add(0,templist)方法来实现倒序

算法如下：

/**
* 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>> resultlist = new ArrayList<>();
if(root==null)
return resultlist;
List<Integer> templist = new ArrayList<Integer>();
LinkedList<TreeNode> s = new LinkedList<TreeNode>();
s.add(root);
int curIndex = 1;
int nextIndex = 0;
while(!s.isEmpty()){
TreeNode node = s.poll();
curIndex--;
templist.add(node.val);
if(node.left!=null){
s.add(node.left);
nextIndex++;
}
if(node.right!=null){
s.add(node.right);
nextIndex++;
}
if(curIndex==0){
resultlist.add(0,templist);
templist = new ArrayList<Integer>();
curIndex = nextIndex ;
nextIndex = 0;
}
}
return resultlist;

}
}