数据结构与算法中二叉树子结构的详解

时间:2021-10-23 07:50:25

数据结构与算法中二叉树子结构的详解

需求

输入两棵二叉树A,B,判断B是不是A的子结构。(ps:我们约定空树不是任意一个树的子结构)

树的描述:

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class TreeNode {
  int val = 0;
  TreeNode left = null;
  TreeNode right = null;
 
  public TreeNode(int val) {
    this.val = val;
 
  }
}

解决思路

使用了栈将元素入栈,并不断的弹出元素,弹出一个元素的时候,拼接成字符串,并用特殊符号进行区分,该方法主要是按照先序遍历的方式将树节点的数据信息拼接为字符串,这样,两个树的节点拼接而成的串进行判断是不是包含。

不过,有的资料上说可以通过递归的方式进行,但是我感觉以及实践以后发现是错误的。后面会给出代码,读者自行尝试。

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public static boolean HasSubtree2(TreeNode root1, TreeNode root2) {
 
    if (root2 == null)
      return false;
    String str = "";
    Stack<TreeNode> stack = new Stack<TreeNode>();
    stack.push(null);
    stack.push(root1);
    TreeNode node = null;
    while ((node = stack.pop()) != null) {
      str += '_' + node.val + '_';
 
      if (node.right != null) {
        stack.push(node.right);
      }
      if (node.left != null) {
        stack.push(node.left);
      }
    }
 
    String str2 = "";
    node = null;
    stack.push(null);
    stack.push(root2);
    while ((node = stack.pop()) != null) {
      str2 += '_' + node.val + '_';
 
      if (node.right != null) {
        stack.push(node.right);
      }
      if (node.left != null) {
        stack.push(node.left);
      }
    }
 
    if (str.contains(str2)) {
      return true;
    } else {
      return false;
    }
  }

树的构建

二叉树而言,可以通过数组的方式进行存放,首节点放在数组0号位置处,其左节点在1号位置处,其右节点在2号位置处。由此该index的映射关系为:

index_parent.left => 2* index_parent + 1;
index_parent.right=> 2* index_parent + 2;

构建思路,左节点和右节点分别构建,根节点的左节点就一直追溯其子节点,根节点的右节点一直追溯其子节点,由此,形成的是递归的结构。

代码如下:

注:这里数组中通过-1作为区分,读者可自行扩充。

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public static TreeNode getTree(int[] node, int index) {
 
    if (index >= node.length)
      return null;
    TreeNode n = null;
    if (node[index] != -1) {
      n = new TreeNode(node[index]);
      n.left = getTree(node, index * 2 + 1);
      n.right = getTree(node, index * 2 + 2);
    }
    return n;
  }

完整代码

包括了资料中提供的代码,但是经过测试如下用例中是错误的,但是理论上说tree2应该是tree1的子结构才对。

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import java.util.Stack;
 
public class HasSubtree {
 
  public static void main(String[] args) {
 
    TreeNode tree = getTree(new int[] { 8, 8, 7, 9, 2, -1, -1, -1, -1, 4, 7 }, 0);
    TreeNode tree2 = getTree(new int[] { 2, 4, 7 }, 0);
    boolean bool = HasSubtree(tree, tree2);
    System.out.println(bool);
 
    boolean bool2 = HasSubtree2(tree, tree2);
    System.out.println(bool2);
  }
 
  public static boolean HasSubtree2(TreeNode root1, TreeNode root2) {
 
    if (root2 == null)
      return false;
    String str = "";
    Stack<TreeNode> stack = new Stack<TreeNode>();
    stack.push(null);
    stack.push(root1);
    TreeNode node = null;
    while ((node = stack.pop()) != null) {
      str += '_' + node.val + '_';
 
      if (node.right != null) {
        stack.push(node.right);
      }
      if (node.left != null) {
        stack.push(node.left);
      }
    }
 
    String str2 = "";
    node = null;
    stack.push(null);
    stack.push(root2);
    while ((node = stack.pop()) != null) {
      str2 += '_' + node.val + '_';
 
      if (node.right != null) {
        stack.push(node.right);
      }
      if (node.left != null) {
        stack.push(node.left);
      }
    }
 
    if (str.contains(str2)) {
      return true;
    } else {
      return false;
    }
  }
 
  public static TreeNode getTree(int[] node, int index) {
 
    if (index >= node.length)
      return null;
    TreeNode n = null;
    if (node[index] != -1) {
      n = new TreeNode(node[index]);
      n.left = getTree(node, index * 2 + 1);
      n.right = getTree(node, index * 2 + 2);
    }
    return n;
  }
 
  public static boolean HasSubtree(TreeNode root1, TreeNode root2) {
 
    boolean result = false;
    if (root1 != null && root2 != null) {
 
      if (root1.val == root2.val) {
        result = isSubTree(root1, root2);
      }
 
      if (!result) {
        result = isSubTree(root1.left, root2);
      }
 
      if (!result) {
        result = isSubTree(root1.right, root2);
      }
    }
 
    return result;
  }
 
  private static boolean isSubTree(TreeNode root1, TreeNode root2) {
 
    if (root1 == null)
      return false;
    if (root2 == null)
      return true;
    if (root1.val != root2.val)
      return false;
 
    return isSubTree(root1.left, root2.left)
        && isSubTree(root1.right, root2.right);
  }
 
}
 
class TreeNode {
  int val = 0;
  TreeNode left = null;
  TreeNode right = null;
 
  public TreeNode(int val) {
    this.val = val;
 
  }
 
}

感谢阅读,希望能帮助到大家,谢谢大家对本站的支持!

原文链接:http://blog.csdn.net/wangyang1354/article/details/70767779