LeetCode OJ：Binary Tree Level Order Traversal II（二叉树的层序遍历）

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]
]

依旧是层序遍历而已，更另一篇博客一样，bfs或者dfs之后reverse一下就可以了，这里给出bfs版本，dfs见另一篇博文：

 /**
  * Definition for a binary tree node.
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 private:
     struct Node{
         TreeNode * treeNode;
         int level;
         Node(){}
         Node(TreeNode * nd, int lv)
         :treeNode(nd), level(lv){}
     };
 public:
     vector<vector<int>> levelOrderBottom(TreeNode* root) {
         vector<vector<int>> ret;
         if(!root)
             return ret;
         queue<Node> nodeQueue;
         nodeQueue.push(Node(root, ));
         int dep = -;
         while(!nodeQueue.empty()){
             Node node = nodeQueue.front();
             if(node.treeNode->left)
                 nodeQueue.push(Node(node.treeNode->left, node.level + ));
             if(node.treeNode->right)
                 nodeQueue.push(Node(node.treeNode->right, node.level + ));
             if(dep == node.level)
                 ret[dep].push_back(node.treeNode->val);
             else{
                 vector<int> tmp;
                 dep++;
                 ret.push_back(tmp);
                 ret[dep].push_back(node.treeNode->val);
             }
             nodeQueue.pop();    //不要忘了
         }
         reverse(ret.begin(), ret.end());
         return ret;
     }
 };

上面写的可能结果是对的，但是好像不是题目本来的意思，下面用dfs重新写一遍

 public class Solution {
     public List<List<Integer>> levelOrderBottom(TreeNode root) {
         List<List<Integer>> ret = new ArrayList<List<Integer>>();
         dfs(ret, root, 0);
         Collections.reverse(ret);
         return ret;
     }

     public void dfs(List<List<Integer>> ret, TreeNode root, int dep){
         if(root == null)
             return;
         if(ret.size() <= dep){
             List<Integer> tmp = new ArrayList<Integer>();
             ret.add(tmp);
         }
         ret.get(dep).add(root.val);
         if(root.left != null) dfs(ret, root.left, dep+1);
         if(root.right != null) dfs(ret, root.right, dep+1);
     }
 }