[LeetCode] 107. Binary Tree Level Order Traversal II Java

標籤：efi   tom   values   init   binary   style   tree node   erb   bsp   

題目：

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]]

 題意及分析：給出一顆二叉樹廣度遍曆的結果，從分葉節點到根節點。和前面根節點到分葉節點類似，只是結果用Collections.reverse反轉一下即可。

代碼：

/** * 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>> res = new ArrayList<>();        if(root==null) return res;        List<Integer> list = new ArrayList<>();        Queue<TreeNode> queue=new LinkedList<>();        queue.offer(root);        Queue<TreeNode> queue1=new LinkedList<>();        while(!queue.isEmpty()||!queue1.isEmpty()){            queue1.clear();            while(!queue.isEmpty()){        //遍曆一層                TreeNode now = queue.poll();                list.add(now.val);                if(now.left!=null)                    queue1.offer(now.left);                if(now.right!=null)                    queue1.offer(now.right);            }            queue=new LinkedList<>(queue1);       //進行下一層的遍曆            res.add(new ArrayList<>(list));            list.clear();       //清空        }        Collections.reverse(res);        return res;    }}

 

 

 

 

[LeetCode] 107. Binary Tree Level Order Traversal II Java