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]
]
难度：easy
题目：给定二叉树，返回由底到上的层次遍历值。
思路：层次遍历
Runtime: 1 ms, faster than 97.69% of Java online submissions for Binary Tree Level Order Traversal II.Memory Usage: 26.4 MB, less than 80.95% of Java online submissions for Binary Tree Level Order Traversal II.
/**
* 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) {
LinkedList<List<Integer>> result = new LinkedList<>();
if (null == root) {
return result;
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
int levelCnt = 1;
while (!queue.isEmpty()) {
int cnt = 0;
List<Integer> llt = new ArrayList<>();
for (int i = 0; i < levelCnt; i++) {
TreeNode node = queue.poll();
llt.add(node.val);
if (node.left != null) {
queue.add(node.left);
cnt++;
}
if (node.right != null) {
queue.add(node.right);
cnt++;
}
}
levelCnt = cnt;
result.addFirst(llt);
}
return result;
}
}