给定一个二叉树,返回其节点值自底向上的层次遍历。 (即按从叶节点所在层到根节点所在的层,逐层从左向右遍历)
例如:
给定二叉树 [3,9,20,null,null,15,7]
,
3 / \ 9 20 / \ 15 7
返回其自自底向上的层次遍历为:
[ [15,7], [9,20], [3] ]
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] ]
个人思路:
参考二叉树的层次遍历的思路,但总的结果用栈存储,最后再一次弹出,从而使各个分数组逆序排列
代码(JavaScript):
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
Queue<TreeNode> queue=new LinkedList<TreeNode>();
Stack<List<Integer>> visited=new Stack<List<Integer>>();
List<List<Integer>> result=new LinkedList<List<Integer>>();
int row=0,col=0;
if(root!=null){queue.offer(root);}
else{return visited;}
while(!queue.isEmpty()){
int level=queue.size();
List<Integer> sub_visited=new LinkedList<Integer>();
for(int i=0;i<level;i++){
if(queue.peek().left!=null){queue.offer(queue.peek().left);}
if(queue.peek().right!=null){queue.offer(queue.peek().right);}
sub_visited.add(queue.poll().val);
}
visited.push(sub_visited);
}
while(!visited.isEmpty()){
result.add(visited.pop());
}
return result;
}
}