【LeetCode】Binary Tree Level Order Traversal II

Binary Tree Level Order Traversal II

Total Accepted: 4229 Total Submissions: 13624 My Submissions

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.

这道题和Binary Tree Level Order
Traversal一样，只不过这个结果需要逆序一下。

也就多了几行代码而已。

ArrayList<ArrayList<Integer>> newlist = new ArrayList<ArrayList<Integer>>();  
    int size = list.size();
    for(int i = size-1; i >= 0 ; i--){
        newlist.add(list.get(i));
    }

还是BFS吧，没什么好说的。
Java AC

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
        ArrayList<ArrayList<Integer>> list = new ArrayList<ArrayList<Integer>>();  
        if(root == null){  
            return list;  
        }  
        bfs(root,list); 
        ArrayList<ArrayList<Integer>> newlist = new ArrayList<ArrayList<Integer>>();  
        int size = list.size();
        for(int i = size-1; i >= 0 ; i--){
            newlist.add(list.get(i));
        }
        return newlist;  
    }
    public void bfs(TreeNode root,ArrayList<ArrayList<Integer>> list){  
        Queue<TreeNode> queue = new LinkedList<TreeNode>();  
        queue.offer(root);  
        while(!queue.isEmpty()){  
            int size = queue.size();  
            ArrayList<Integer> tempList = new ArrayList<Integer>();  
            for(int i = 0; i < size; i++){  
                TreeNode point = queue.peek();  
                queue.poll();  
                tempList.add(point.val);  
                if(point.left != null){  
                    queue.offer(point.left);  
                }  
                if(point.right != null){  
                    queue.offer(point.right);  
                }  
            }  
            list.add(tempList);  
        }  
    }  
}