• <ins id="pjuwb"></ins>
    <blockquote id="pjuwb"><pre id="pjuwb"></pre></blockquote>
    <noscript id="pjuwb"></noscript>
          <sup id="pjuwb"><pre id="pjuwb"></pre></sup>
            <dd id="pjuwb"></dd>
            <abbr id="pjuwb"></abbr>
            posts - 195,  comments - 30,  trackbacks - 0
            Leaf Nodes
            Status In/Out TIME Limit MEMORY Limit Submit Times Solved Users JUDGE TYPE
            stdin/stdout 1s 10240K 222 102 Standard

            Kate is learning about binary tree. She think she know everything you know about binary trees. Wait, you don't know binary tree? Find a book about data structures, and it will just take you about three minutes. Now here is a binary tree:

                                                3
            / \
            /   \
            2     4
            / \     \
            /   \     \
            0     1     6
            /
            /
            5
            

            Kate think she also know something that you may not notice. First, for some type of binary trees, only the leaf nodes have the meaning (leaf node is the node which has no sub trees, for the tree above, the leaf nodes are 0 1 5), an example is the Huffman Tree. Second, she guess that if you know the preorder traversal and the postorder traversal of a binary tree, you can ensure the leaf node of the tree, and their order.

            For the tree above, the preorder travesal is 3 2 0 1 4 6 5 and the postorder travesal is 0 1 2 5 6 4 3, the leaf nodes in order(from left to right) are 0 1 5.

            But now the problem is she just guess it, if you can find a way to print a tree's leaf nodes in order using its preorder traversal and postorder traversal, you can say "she is right!"

            Input Specification

            The input file will contain one or more test cases. In each test case there is a integer n (0<=n <= 10000), indicating the tree have n nodes, then followed two lists of integers, either contains n integers in the range of 0 and n-1 (both included), the first list is the preorder traversal, and the other is the postorder traversal.

            Input is terminated by an interger -1;

            Output Specification

            For each test case, print the tree's leaf nodes in order, each in a line.

            Sample Input

            7
            3 2 0 1 4 6 5
            0 1 2 5 6 4 3
            -1

            Sample Output

            0
            1
            5
            根據一個重要結論,無論是先根還是后根遍歷,左子樹的結點總是出現在右子樹結點的前面

                                             G  
                                            /   \  
                                          F        B  
                                        /        /   \  
                                      K         C      H  
                                    /   \             /        
                                  D       E         J  
                                    \  
                                      A  
                                    /  
                                  I  
               
              不論先根后根,左子樹的結點總是出現在右子樹結點的前面。  
              G為根樹,先根次序時G后跟F,后根次序時F前有DIAEKF,故DIAEKF為G的左子樹的結點,
              CJHB為G的右子樹的結點。且左右子樹的先根序為:FKDIAE,BCHJ。
              遞歸處理兩子樹即可搞定

            void  find(int preb,int pree,int postb,int poste) 
               {
               int i=s(pre,post[poste-1]);
              int j=s(post,pre[preb+1]);
            //添加處理的代碼
             //判斷是否有左/右支

                find(preb+1,i-1,postb,j);
              find(i,pree,j+1,poste-1);
               }
            但是上面的思路是錯誤的!!!!!!!!!!!!!!
            只知道先序和后序不能能推出樹來  
               
              只有中序和先序或者中序和后序才可以  
               
              不然只知道根節點,但是哪些是左子樹哪些是右子樹就不知道了
            比如先序時1234  
              后序是4321的二叉樹有8種比如:  
                  1                     1              
                  \                   /  
                    2               2  
                /                 /  
              3                 3  
                \             /  
                  4         4

            正確思路:先遍歷后根次序,第一個一定為葉子,設為當前結點,然后依次檢測,如果該點在先根序列中位于當前節點的后面,則為葉子節點,同時更新當前結點。
            posted on 2009-07-14 22:14 luis 閱讀(277) 評論(0)  編輯 收藏 引用 所屬分類: 圖論*矩陣
            <2009年7月>
            2829301234
            567891011
            12131415161718
            19202122232425
            2627282930311
            2345678

            常用鏈接

            留言簿(3)

            隨筆分類

            隨筆檔案

            文章分類

            文章檔案

            友情鏈接

            搜索

            •  

            最新評論

            閱讀排行榜

            評論排行榜

            99久久人人爽亚洲精品美女| 久久国产成人午夜AV影院| 91久久精品电影| 久久WWW免费人成—看片| 亚洲欧美另类日本久久国产真实乱对白| 女同久久| 丰满少妇人妻久久久久久| 亚洲性久久久影院| 久久精品国产亚洲AV嫖农村妇女| 久久91综合国产91久久精品| 久久涩综合| 狠色狠色狠狠色综合久久 | 久久久亚洲裙底偷窥综合| 久久精品国产亚洲av影院| 久久久久亚洲精品男人的天堂| 久久香蕉国产线看观看精品yw| 精品久久久久久无码免费| 囯产精品久久久久久久久蜜桃| 品成人欧美大片久久国产欧美| 97精品国产97久久久久久免费| 久久久久久毛片免费看| 成人国内精品久久久久影院| 久久精品免费一区二区| 国产精品热久久无码av| 久久综合九色综合精品| 国产精品国色综合久久| 色婷婷综合久久久中文字幕 | 久久精品国产亚洲77777| 久久精品国产免费观看三人同眠| 久久se精品一区二区影院| 久久亚洲国产午夜精品理论片| 久久亚洲精品无码AV红樱桃| 久久婷婷五月综合国产尤物app| 性高湖久久久久久久久AAAAA| 99久久伊人精品综合观看| 青青青青久久精品国产 | 国内精品欧美久久精品| 香蕉久久夜色精品国产小说| 丰满少妇人妻久久久久久| 精品综合久久久久久97超人| 狠狠色婷婷综合天天久久丁香 |