**《劍指offer》17：樹的子結構

題目描述

輸入兩棵二叉樹A，B，判斷B是不是A的子結構。（ps：我們約定空樹不是任意一個樹的子結構）

C++實現：

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    bool doeshavetree(TreeNode* p1, TreeNode* p2)
    {
        if(p2 == NULL)
            return true;
        if(p1 == NULL)
            return false;
        if(p1 -> val != p2 -> val)
            return false;
        
        return doeshavetree(p1 -> left, p2 -> left) && doeshavetree(p1 -> right, p2 -> right);
    }
    
    bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
    {
        bool result = false;
        
        if(!pRoot2 || !pRoot1)
            return false;
        
        if(pRoot1 -> val == pRoot2 -> val)
            result = doeshavetree(pRoot1, pRoot2);
        if(!result)
            result = HasSubtree(pRoot1 -> left, pRoot2);
        if(!result)
            result = HasSubtree(pRoot1 -> right, pRoot2);
        
        return result;
    }
};

python實現：

# -*- 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
        def convert(p):
            if p:
                return str(p.val) + convert(p.left) + convert(p.right)
            else:
                return ''
        if pRoot2:
            if convert(pRoot2) in convert(pRoot1):
                return True
            else:
                return False
        return False