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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] ]
confused what "{1,#,2,3}" means? >
read more on how binary tree is serialized on OJ.
The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.
Here‘s an example:
1
/ 2 3
/
4
5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".思路：还是一样，最后倒置一下就行了
/**
* 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>> ans = new ArrayList<List<Integer>>();
if (root == null) return ans;
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.add(root);
int count = 1, level = 0;
while (!q.isEmpty()) {
List<Integer> tmp = new ArrayList<Integer>();
level = 0;
for (int i = 0; i < count; i++) {
TreeNode t = q.poll();
tmp.add(t.val);
if (t.left != null) {
q.add(t.left);
++level;
}
if (t.right != null) {
q.add(t.right);
++level;
}
}
count = level;
ans.add(tmp);
}
Collections.reverse(ans);
return ans;
}
}
LeetCode Binary Tree Level Order Traversal II
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原文地址：http://blog.csdn.net/u011345136/article/details/45498717
