[Leetcode] Binary tree level order traversal ii二叉树层次遍历

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],
]

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:

   1
  / \
 2   3
    /
   4
    \
     5

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

 

这一题和Binary tree level order traversal的区别在于，输出结果是从下往上一层层的遍历节点，有两种小聪明的方法：一、在上一题的最后反转res然后输出；二、使用栈。得到每一层的节点后，不直接压入res中，先存入栈中，然后利用栈先进后出的特点，再压入res，实现反转。这两题关键的部分在于如何单独得到每一层的节点。以下两种方法的主体部分也可以用上一题中的方法一，其核心思想不变

方法一：最后反转

 /**
  * 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);

         while( !Q.empty())
         {
             int count=;
             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++;
             }
             res.push_back(levNode);
         }
         reverse(res.begin(),res.end());    //在上一题的基础上增加一行
         return res;
     }
 };

方法二：利用栈

/**
 * 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=;
            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);   //压入栈
        }
        while(!stk.empty())       //出栈
        {
            res.push_back(stk.top());
            stk.pop();
        }
        return res;
    }
};