使用后顺序检查子树

Question

我正在尝试编写一个程序来检查树T2是否是树T1中存在的子树。 我通过将树的状态存储到字符串中并检查T2的子字符串是否是T1的子字符串中的子字符串来进行遍历。 由于某种原因，如果我在遍历方法中按顺序执行以生成字符串来保存二进制树的表示形式，则代码将失败。 为什么会这样？ 进行后期订购效果很好。 这是我的程序失败的输入情况：

{9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9 ，9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9} {9， 9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9， 9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9， 9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9， 9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9， 9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9， 9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9， 9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9， 9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9， 9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9， 9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9， 9,9,9}

可以在http://www.lintcode.com/help/binary-tree-representation/中找到二进制树的表示格式。

public class Solution {
/*
 * @param T1: The roots of binary tree T1.
 * @param T2: The roots of binary tree T2.
 * @return: True if T2 is a subtree of T1, or false.
 */
public boolean isSubtree(TreeNode T1, TreeNode T2) {
    // write your code here
    if(T1 == null && T2 == null) return true;
    if(T1 == null) return false;
    if(T2 == null) return true;
    StringBuilder sbT1 = new StringBuilder("");
    StringBuilder sbT2 = new StringBuilder("");
    traverse(T1, sbT1);
    traverse(T2, sbT2);
    return sbT1.toString().indexOf(sbT2.toString()) != -1;
}

public void traverse(TreeNode a, StringBuilder sb) {
    if(a == null) {
        sb.append("POP");
        return;
    } else {
        traverse(a.left, sb);
        sb.append(a.val);
        traverse(a.right, sb);

    }
}

}

Answer 1

您实现的递归算法未按所需顺序提供树的序列化。 想象一下它是如何工作的。 左子树的所有项目将始终在右子树的任何项目之前，因为右子树将始终在左子树完成后进行处理。 在您提到的页面中描述的表单中，所有级别的项目都应彼此跟随，并且在完成给定级别的所有项目之后，应处理下一个级别的项目。 要按照描述的顺序编写树，您需要一个更复杂的算法。

而且，所描述的将树表示为字符串的方式不允许仅通过搜索子字符串来搜索子树。 考虑下面的树：

     1
   /    \
  2      3
 / \    / \
4   5  6   7

它有一个子树

    3
   / \ 
  6   7 

但是字符串“ 1,2,3,4,5,6,7”不包含“ 3,6,7”。

在树中查找子树的一种更简单，更直接的方法是直接对树进行操作，而无需将它们转换为字符串。

为了稍微发挥我的创造力，我考虑了一种递归算法，该算法应使用所描述的规则将树转换为字符串，并实现了一种简单算法来直接搜索子树，因此您可以确保

• 字符串形式的树的子树不是整个树的相应字符串的子字符串；
• 在树中搜索子树比将树转换为字符串要简单得多，即使转换算法允许在结果字符串中搜索子字符串：

---

import java.util.ArrayList;
import java.util.Arrays;

public class So_47736214 {

  static String[][] testData = {
    {"1,2,3,4,5,#,6,#,#,7,8", "4,5,6"},                 // Does not contain
    {"1,2,3,4,5,#,6,#,#,7,8", "1,2,3,4,5,#,6,#,#,7,8"}, // Are equal 
    {"1,2,3,4,5,#,6,#,#,7,8", "2,4,5"},         // Contains in the left subtree   
    {"1,2,3,4,5,#,6,#,#,7,8", "3,#,6,#,7"},     // Contains in the right subtree 
    {"7,8,9", "10,11,12"},                      // Have nothing in common       
  };


  static class TreeItem {
    int value; TreeItem left; TreeItem right;
    TreeItem(int v) { this.value = v; }
  }

  static TreeItem tree1, tree2; // To hold the two trees to be compared

  /** Converts a string to a tree */
  private static TreeItem deserialize(String s) { 
    // First, prepare the layers of the tree as AttayLists containing items with values
    String[] ss = s.split(","); if (ss.length < 2) return null; // a separate string for each item 
    int pos = 0, len = 1;                       // Start position of the next layer and its length
    ArrayList<ArrayList<TreeItem>> layers = new ArrayList<>(); // A structure to hold the tree layers
    while (pos < ss.length) {
      String[] layerStrings = Arrays.copyOfRange(ss, pos, pos + len); // Strings for the next layer 
      ArrayList<TreeItem> layerItems = new ArrayList<>();             // Container for the layer's items  
      for (int i = 0; i < layerStrings.length; i++) 
        layerItems.add( ( layerStrings[i] == null || layerStrings[i].equals("#") ) ? 
          null : new TreeItem(Integer.parseInt(layerStrings[i])));    // Fill with items or nulls 
      layers.add(layerItems);
      pos += len; len *= 2;       // Start pos and lengths of the next layer
    }
    // Now the 2-d array of items is built. Establish links between an item and its subtrees 
    ArrayList<TreeItem> prevLayer = layers.get(0);  
    for (int i = 1; i < layers.size(); i++) { // Start from 1 -- skip layer 0  
      ArrayList<TreeItem> layer = layers.get(i);
      for (int j = 0; j < prevLayer.size(); j++ ) {
        TreeItem prevLayerItem = prevLayer.get(j);
        if (prevLayerItem != null) {
          prevLayerItem.left = layer.get(j * 2);      // left subtree is a next layer's item 
          prevLayerItem.right = layer.get(j * 2 + 1); // right subtree
        }
      };
      prevLayer = layer;
    }
    return layers.get(0).get(0); // The tree is ready. It's the root
  }

  private static void deserialize(String s1, String s2) {
    tree1 = deserialize(s1); 
    tree2 = deserialize(s2); 
  }

  static int levelCount;

  /** Puts the item into the appropriate position in the list of the layers 
   * and recursively processes the item's subtrees */  
  private static void insertItem(TreeItem item, ArrayList<TreeItem[]> layerList, int idx, int level) {
    TreeItem[] levelItems;
    if (layerList.size() < level + 1)  // Needs to create a new array for the level
      layerList.add(levelItems = new TreeItem[1 << level]); // size of the array = pow(2, level)  
    else                            
      levelItems = layerList.get(level);
    levelItems[idx] = item;
    idx *= 2; level++;
    if (item.left != null)
      insertItem(item.left, layerList, idx, level);
    if (item.right != null)
      insertItem(item.right, layerList, idx+ 1, level);
  }

  /** Writes the values from the list of layers into a string */ 
  private static String arraysToString(ArrayList<TreeItem[]> layerList) {
    StringBuilder sb = new StringBuilder();
    int strLen = 0;  
    for (TreeItem[] layer: layerList) 
      for (int i = 0; i < layer.length; i++) 
        if (layer[i] != null) {
          sb.append(layer[i].value + ",");
          strLen = sb.length();
        } else 
          sb.append("#,");
    return sb.length() == 0? "" : sb.substring(0, strLen - 1);
  }

  /** Converts a tree into a string */
  private static String serialize(TreeItem tree) {
    ArrayList<TreeItem[]> items = new ArrayList<>();
    levelCount = 0;
    insertItem(tree, items, 0, 0);
    return arraysToString(items); 
  }

  /** Converts a pair of trees into strings and look for substrings */  
  private static void test1(TreeItem t1, TreeItem t2) {
    System.out.println("Test1:");
    String s1 = serialize(t1), s2 = serialize(t2);
    String result = s1.contains(s2)? "contains" : "doesn't contain"; 
    System.out.format("  '%s' %s '%s'\n", s1, result, s2);
  }

  /** Checks if the trees are equal */
  private static boolean areEqual(TreeItem t1, TreeItem t2) {
    if (t1 == null && t2 == null) return true;
    if (t1 == null || t2 == null) return false; 
    return (areEqual(t1.left, t2.left) && areEqual(t1.right, t2.right)); 
  }

  /** Checks if t2 is a subtree of t1 */  
  private static boolean contains(TreeItem t1, TreeItem t2) {
    if (t1 == null && t2 == null) return true;
    if (t1 == null) return false; 
    if (t2 == null) return true; 
    if ( t1.value == t2.value 
          && areEqual(t1.left, t2.left)
          && areEqual(t1.left, t2.left)) 
      return true;
    return (contains(t1.left, t2) || contains(t1.right, t2));
  }

  /** Checks if t1 is a subtree of t1, and shows the result */ 
  private static void test2(TreeItem t1, TreeItem t2) {
    System.out.println("Test2:");
    String s1 = serialize(t1), s2 = serialize(t2);
    String result = (contains(t1, t2)? "contains" : "doesn't contain"); 
    System.out.format("  '%s' %s '%s'\n", s1, result, s2);
  }

  public static void main(String[] args) {
    // runs the two tests through the test dataset
    for (int i = 0; i < testData.length; i++) {
      deserialize(testData[i][0], testData[i][1]); // converts strings into trees  
      test1(tree1, tree2);      // looks if t1 is a subtree of t1, using string representations  
      test2(tree1, tree2);      // looks if t1 is a subtree of t1, directly with trees
      System.out.println("");
    }
  }

}