剑指offer66题--Java实现，c++实现和python实现 17.树的子结构

题目描述

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）

C++

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
    {
        if (pRoot1 == nullptr || pRoot2 == nullptr)
            return false;   //直接返回错误
        return isTheSameTree(pRoot1, pRoot2) || 
            HasSubtree(pRoot1->left, pRoot2) || HasSubtree(pRoot1->right, pRoot2);
    }
    bool isTheSameTree(TreeNode* pRoot1, TreeNode* pRoot2)
    {
        if (pRoot2 == nullptr)
            return true;
        if (pRoot1 == nullptr && pRoot2 != nullptr)
            return false;
        return pRoot1->val == pRoot2->val && 
            isTheSameTree(pRoot1->left, pRoot2->left) && isTheSameTree(pRoot1->right, pRoot2->right);
    }
};

Java

/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
public class Solution {
    public 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 == false)
	                result = HasSubtree(root1.left,root2);
	            if(result == false)
	               result = HasSubtree(root1.right,root2);
	            
	         }
	        return result;
	    }
	    private boolean isSubTree(TreeNode root1,TreeNode root2)
	    {
	        if(root2 == null)
	            return true;
	        if(root1 == null)
	            return false;
	        if(root1.val != root2.val)
	            return false;
	        return isSubTree(root1.left,root2.left)&&isSubTree(root1.right,root2.right);
	        
	    }
}

python 

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    def HasSubtree(self, pRoot1, pRoot2):
        # write code here
        if pRoot1 == None or pRoot2 == None:
            return False
        return self.isSubtree(pRoot1, pRoot2)

    def isSubtree(self, p1, p2):
        if p2 == None:
            return True
        if p1 == None:
            return p1 == p2
        res = False
        if p1.val == p2.val:
            res = self.isSubtree(p1.left, p2.left) and self.isSubtree(p1.right, p2.right)
        return res or self.isSubtree(p1.left, p2) or self.isSubtree(p1.right, p2)