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：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 private:
12     struct Node{
13         TreeNode * treeNode;
14         int level;
15         Node(){}
16         Node(TreeNode * nd, int lv)
17         :treeNode(nd), level(lv){}
18     };
19 public:
20     vector<vector<int>> levelOrderBottom(TreeNode* root) {
21         vector<vector<int>> ret;
22         if(!root)
23             return ret;
24         queue<Node> nodeQueue;
25         nodeQueue.push(Node(root, 0));
26         int dep = -1;
27         while(!nodeQueue.empty()){
28             Node node = nodeQueue.front();
29             if(node.treeNode->left)
30                 nodeQueue.push(Node(node.treeNode->left, node.level + 1));
31             if(node.treeNode->right)
32                 nodeQueue.push(Node(node.treeNode->right, node.level + 1));
33             if(dep == node.level)
34                 ret[dep].push_back(node.treeNode->val);
35             else{
36                 vector<int> tmp;
37                 dep++;
38                 ret.push_back(tmp);
39                 ret[dep].push_back(node.treeNode->val);
40             }
41             nodeQueue.pop();    //不要忘了
42         }
43         reverse(ret.begin(), ret.end());
44         return ret;
45     }
46 };

 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

 1 public class Solution {
 2     public List<List<Integer>> levelOrderBottom(TreeNode root) {
 3         List<List<Integer>> ret = new ArrayList<List<Integer>>();
 4         dfs(ret, root, 0);
 5         Collections.reverse(ret);
 6         return ret;
 7     }
 8     
 9     public void dfs(List<List<Integer>> ret, TreeNode root, int dep){
10         if(root == null)
11             return;
12         if(ret.size() <= dep){
13             List<Integer> tmp = new ArrayList<Integer>();
14             ret.add(tmp);
15         }
16         ret.get(dep).add(root.val);
17         if(root.left != null) dfs(ret, root.left, dep+1);
18         if(root.right != null) dfs(ret, root.right, dep+1);
19     }
20 }

 

posted @ 2015-10-23 21:30  eversliver  阅读(236)  评论(0)  编辑  收藏  举报