Binary Tree Level Order Traversal II

标签：binarys   queue   list   collections   

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]
]

与Binary Tree Level Order Traversal完全一致 只要最后将Binary Tree Level Order Traversal的结果翻转一下 可以利用Collections的reverse函数 代码如下：

public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<Integer> tmp=new ArrayList<Integer>();
			List<List<Integer>>	res=new ArrayList<List<Integer>>();
			if(root==null) return res;
			int count=1;
			int level=0;
			Queue<TreeNode> que =new LinkedList<TreeNode>();
			que.offer(root);
			while(que.isEmpty()!=true){
				tmp.clear();
				level=0;
				for(int i=0;i<count;i++){
					root=que.peek();
					tmp.add(root.val);
					que.poll();
					if(root.left!=null){
						que.offer(root.left);
						level++;
					}
					if(root.right!=null){
						que.offer(root.right);
						level++;
					}
				}
				List<Integer> tmpa=new ArrayList<Integer>();
				tmpa.addAll(tmp);
				count=level;
				res.add(tmpa);		
			}
			Collections.reverse(res);
			return res;
    }
}

Binary Tree Level Order Traversal II

标签：binarys   queue   list   collections   

原文地址：http://blog.csdn.net/u012734829/article/details/42266053