【LeetCode】Binary Tree Level Order Traversal II

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题目

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,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

[  [15,7],  [9,20],  [3]]

解答

和Binary Tree Level Order Traversal类似，只是将它的结果翻转而已，代码如下：

/** * Definition for binary tree * 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>> ret=new ArrayList<List<Integer>>();        dfs(root,0,ret);        List<List<Integer>> result=new ArrayList<List<Integer>>();        //将List翻转        for(int i=ret.size()-1;i>=0;i--){            result.add(ret.get(i));        }        return result;    }    //分层遍历-递归    private static void dfs(TreeNode root,int level,List<List<Integer>> ret){        if(root==null){            return;        }        //添加一个新的ArrayL表示新的一层if(level>=ret.size()){ret.add(new ArrayList<Integer>());}ret.get(level).add(root.val);  //把节点添加到表示那一层的ArrayList里dfs(root.left,level+1,ret);    //递归处理下一层的左子树和右子树        dfs(root.right,level+1,ret);    }}//迭代法：public List<List<Integer>> levelOrderBottom(TreeNode root){List<List<Integer>> result=new ArrayList<List<Integer>>();if(root==null){return result;}Queue<TreeNode> queue=new LinkedList<TreeNode>();queue.add(root);List<List<Integer>> list=new ArrayList<List<Integer>>();List<Integer> item=new ArrayList<Integer>();int currentLevel=1;int nextLevel=0;while(!queue.isEmpty()){TreeNode cur=queue.remove();currentLevel--;item.add(cur.val);if(cur.left!=null){queue.add(cur.left);nextLevel++;}if(cur.right!=null){queue.add(cur.right);nextLevel++;}if(currentLevel==0){list.add(item);item=new ArrayList<Integer>();currentLevel=nextLevel;nextLevel=0;}}//翻转listfor(int i=list.size()-1;i>=0;i--){result.add(list.get(i));}return result;}

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