【題目】

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]
]

【程式碼】

/*********************************
*   日期：2014-10-16
*   作者：SJF0115
*   題號: Binary Tree Level Order Traversal II 
*   來源：https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/
*   結果：AC
*   來源：LeetCode
*   總結：
**********************************/
#include <iostream>
#include <malloc.h>
#include <vector>
#include <queue>
using namespace std;

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> > vec;
        vector<int> v;
        if(root == NULL){
            return vec;
        }
        queue<TreeNode*> q;
        q.push(root);
        //當前層的個數
        int count = 1;
        //下一層的個數
        int nextCount = 0;
        while(!q.empty()){
            //取佇列頭元素
            TreeNode* p = q.front();
            //儲存元素
            v.push_back(p->val);
            q.pop();
            count--;
            //左右子節點
            if(p->left){
                q.push(p->left);
                nextCount++;
            }
            if(p->right){
                q.push(p->right);
                nextCount++;
            }
            //一層訪問完畢
            if(count == 0){
                count = nextCount;
                nextCount = 0;
                vec.insert(vec.begin(),v);
                v.clear();
            }
        }
        return vec;
    }
};

//按先序序列建立二叉樹
int CreateBTree(TreeNode* &T){
    char data;
    //按先序次序輸入二叉樹中結點的值（一個字元），‘#’表示空樹
    cin>>data;
    if(data == '#'){
        T = NULL;
    }
    else{
        T = (TreeNode*)malloc(sizeof(TreeNode));
        //生成根結點
        T->val = data-'0';
        //構造左子樹
        CreateBTree(T->left);
        //構造右子樹
        CreateBTree(T->right);
    }
    return 0;
}

int main() {
    Solution solution;
    TreeNode* root(0);
    CreateBTree(root);
    vector<vector<int> > v = solution.levelOrderBottom(root);
    for(int i = 0;i < v.size();i++){
        for(int j = 0;j < v[i].size();j++){
            cout<<v[i][j];
        }
        cout<<endl;
    }
}
//13#5##49###