和之前一题很像,把之前一题的解法答案reverse即可
Java
/**
* 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>> levelOrder(TreeNode root) {
List<List<Integer>> result=new LinkedList<List<Integer>>();
count(result,root,0);
Collections.reverse(result);
return result;
}
public void count(List<List<Integer>> list,TreeNode root,int level)
{
if(root==null) return;
if(level>=list.size())
{
list.add(0,new LinkedList<Integer>());
}
count(list,root.left,level+1);
count(list,root.right,level+1);
list.get(list.size()-level-1).add(root.val);
}
}
Java,貌似不需要reverse,递归的时候list中加值的顺序可以改
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
levelHelper(res, root, 0);
return res;
}
public void levelHelper(List<List<Integer>> res, TreeNode root, int height) {
if (root == null) return;
if (height >= res.size()) {
res.add(new LinkedList<Integer>());
}
res.get(height).add(root.val);
levelHelper(res, root.left, height+1);
levelHelper(res, root.right, height+1);
}