11, tree

Subject description:

Two binary inputs A, B, B is judged not substructure A's. (Ps: we agreed empty tree is not a tree any substructure)

Outline of Solution: A, to traverse the tree to the preamble B, if A is B comprises, is not contrary substructure

 public boolean HasSubtree(TreeNode root1,TreeNode root2) {
        if(root1==null || root2==null)
		   return false;
	   StringBuilder sb = new StringBuilder();
	   StringBuilder sb2 = new StringBuilder();
	   String str1 = preOrder(root1,sb);
	   String str2 = preOrder(root2,sb2);
	   if(str1.contains(str2)){
		   return true;
	   }
	 return false;
    }
     public  String  preOrder(TreeNode root,StringBuilder sb){
	   
	   if(root!=null)
		  sb.append(root.val+",");
	   if(root.left!=null){
		   preOrder(root.left,sb);
	   }
	   if(root.right!=null){
		   preOrder(root.right,sb);
	   }
	   return ","+sb.toString();
	   
   }

　　Problem-solving ideas II:

/*思路：参考剑指offer

1、首先设置标志位result = false，因为一旦匹配成功result就设为true，

剩下的代码不会执行，如果匹配不成功，默认返回false

2、递归思想，如果根节点相同则递归调用DoesTree1HaveTree2（），

如果根节点不相同，则判断tree1的左子树和tree2是否相同，

再判断右子树和tree2是否相同

3、注意null的条件，HasSubTree中，如果两棵树都不为空才进行判断，

DoesTree1HasTree2中，如果Tree2为空，则说明第二棵树遍历完了，即匹配成功，

tree1为空有两种情况（1）如果tree1为空&&tree2不为空说明不匹配，

（2）如果tree1为空，tree2为空，说明匹配。

 

*/

 

public class Solution {

     public boolean HasSubtree(TreeNode root1,TreeNode root2) {

         boolean result = false ;

             if (root1 != null && root2 != null ){

                 if (root1.val == root2.val){

                     result = DoesTree1HaveTree2(root1,root2);

                 }

                 if (!result){result = HasSubtree(root1.left, root2);}

                 if (!result){result = HasSubtree(root1.right, root2);}

             }

             return result;

     }

     public boolean DoesTree1HaveTree2(TreeNode root1,TreeNode root2){

             if (root1 == null && root2 != null ) return false ;

             if (root2 == null ) return true ;

             if (root1.val != root2.val) return false ;

             return DoesTree1HaveTree2(root1.left, root2.left) && DoesTree1HaveTree2(root1.right, root2.right);

         }

}

 

 

Subject description:

 Operation of a given binary tree, the binary tree is converted into a source image.

 Problem-solving ideas:

 Tree left subtree and right subtree recursively exchange

 

 public void Mirror(TreeNode root) {
        if(root!=null){
            TreeNode temp=null;
            temp=root.left;
            root.left=root.right;
            root.right=temp;
            if(root.left!=null){
                Mirror(root.left);
            }
            if(root.right!=null){
                Mirror(root.right);
            }
            
        }
    }