[Leetcode] Binary tree level order traversal ii二叉樹層次遍歷

本文轉載自查看原文 2017-06-08 14:29 1744 Leetcode+題

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},

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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

這一題和 Binary tree level order traversal的區別在於，輸出結果是從下往上一層層的遍歷節點，有兩種小聰明的方法：一、在上一題的最后反轉res然后輸出；二、使用棧。得到每一層的節點后，不直接壓入res中，先存入棧中，然后利用棧先進后出的特點，再壓入res，實現反轉。這兩題關鍵的部分在於如何單獨得到每一層的節點。以下兩種方法的主體部分也可以用上一題中的方法一，其核心思想不變

方法一：最后反轉

 1 /**
 2  * Definition for binary tree
 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 {
12 public:
13     vector<vector<int>> levelOrderBottom(TreeNode* root)
14     {
15         vector<vector<int>> res;
16         queue<TreeNode *> Q;
17         if(root)    Q.push(root);
18 
19         while( !Q.empty())
20         {
21             int count=0;
22             int levCount=Q.size();
23             vector<int> levNode;
24 
25             while(count<levCount)
26             {
27                 TreeNode *curNode=Q.front();
28                 Q.pop();
29                 levNode.push_back(curNode->val);
30                 if(curNode->left)
31                     Q.push(curNode->left);
32                 if(curNode->right)
33                     Q.push(curNode->right);
34                 count++;
35             }
36             res.push_back(levNode);
37         }
38         reverse(res.begin(),res.end());    //在上一題的基礎上增加一行
39         return res;
40     }
41 };

方法二：利用棧

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution
{
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root)
    {
        vector<vector<int>> res;
        queue<TreeNode *> Q;
        if(root)    Q.push(root);
        stack<vector<int>> stk;
        while( !Q.empty())
        {
            int count=0;
            int levCount=Q.size();
            vector<int> levNode;

            while(count<levCount)
            {
                TreeNode *curNode=Q.front();
                Q.pop();
                levNode.push_back(curNode->val);
                if(curNode->left)
                    Q.push(curNode->left);
                if(curNode->right)
                    Q.push(curNode->right);
                count++;
            }
            stk.push(levNode);   //壓入棧
        }
        while(!stk.empty())       //出棧
        {
            res.push_back(stk.top());
            stk.pop();
        }
        return res;
    }
};

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