LeetCode——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]
]

原题链接：https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/

题目：给定一个二叉树，从底到顶返回节点的层序遍历的值。（如，从左到右，一层一层）

思路：上一题的结果上面逆转一下即可。完全不用出这题吧？

public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> list = new ArrayList<List<Integer>>();
if (root == null)
return list;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
while (!queue.isEmpty()) {
List<Integer> li = new ArrayList<Integer>();
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
li.add(node.val);
if (node.left != null)
queue.add(node.left);
if (node.right != null)
queue.add(node.right);
}
list.add(li);
}
List<List<Integer>> ret = new ArrayList<List<Integer>>();
for (int i = list.size() - 1; i >= 0; i--) {
ret.add(list.get(i));
}
return ret;
}

// Definition for binary tree
public class TreeNode {
int val;
TreeNode left;
TreeNode right;

TreeNode(int x) {
val = x;
}
}