剑指offer--面试题18：树的子结构

题目描述

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    """
    1.先在pRoot1中找到与pRoot2根节点一样值得节点，找到则转2
    2.递归检查除根节点外的其他节点值是否相等
    """
    def HasSubtree(self, pRoot1, pRoot2):
        # write code here
        #采用先序遍历查找pRoot1
        if pRoot2 is None:
            return False
        result = False
        if pRoot1:
            if pRoot1.val==pRoot2.val:
                result = self.check(pRoot1, pRoot2)
            #if not result:#这种写法也AC
            #    result = self.HasSubtree(pRoot1.left, pRoot2) or self.HasSubtree(pRoot1.right, pRoot2)
            if not result:
                result = self.HasSubtree(pRoot1.left, pRoot2)
            if not result:
                result = self.HasSubtree(pRoot1.right, pRoot2)
             
        return result
     
    def check(self, pRoot1, pRoot2):
        if pRoot2 is None:
            return True##############注意：这里是True，而不是False
        if pRoot1 is None:
            return False
        if pRoot1.val==pRoot2.val:
            return self.check(pRoot1.left, pRoot2.left) and self.check(pRoot1.right, pRoot2.right)
        return False

/*
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(pRoot2==NULL)
            return false;
        bool result = false;
        if(pRoot1){
            result = check(pRoot1, pRoot2);
            if(!result)
                result = HasSubtree(pRoot1->left, pRoot2) || HasSubtree(pRoot1->right, pRoot2);
        }
        return result;
    }
     
    bool check(TreeNode* pRoot1, TreeNode* pRoot2){
        if(pRoot2==NULL)
            return true;
        if(pRoot1==NULL)
            return false;
        if(pRoot1->val==pRoot2->val)
            return check(pRoot1->left, pRoot2->left
4000
) && check(pRoot1->right, pRoot2->right);
        return false;
    }
};