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]
]
一刷
题解：
BFS， add新的list的时候应该insert到头部，那么层较深，就会创建新的头部并且插在头部。list的size得到树的高度h，然后用h-level得到应该插入哪个list.
/**
* 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>> res = new LinkedList<>();
traverse(root, 0, res);
return res;
}
private void traverse(TreeNode root, int level, List<List<Integer>> res){
if(root==null) return;
if(level>=res.size()) res.add(0, new ArrayList<>());
res.get(res.size()-1-level).add(root.val);
traverse(root.left, level+1, res);
traverse(root.right, level+1, res);
}
}
