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] ]
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思路：
承接上题【Binary Tree Level Order Traversal】，将结果的插入方式改为“头插”即可。
java code:
/**
* 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) {
Queue<TreeNode> queue = new LinkedList<TreeNode>();
List<List<Integer>> ans = new LinkedList<List<Integer>>();
if(root == null) return ans;
queue.offer(root);
while(!queue.isEmpty()) {
int size = queue.size();
List<Integer> subAns = new LinkedList<Integer>();
for(int i=0;i<size;i++) {
TreeNode tmp = queue.poll();
subAns.add(tmp.val);
if(tmp.left != null) queue.offer(tmp.left);
if(tmp.right != null) queue.offer(tmp.right);
}
ans.add(0, subAns); //头插
}
return ans;
}
}LeetCode:Binary Tree Level Order Traversal II
原文：http://blog.csdn.net/itismelzp/article/details/51680040