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  • 100. Same Tree

    1. 问题描述

    Given two binary trees, write a function to check if they are equal or not.

    Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

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    Tags: Tree Depth-first Search
    /**
    * Definition for a binary tree node.
    * struct TreeNode {
    * int val;
    * TreeNode *left;
    * TreeNode *right;
    * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
    * };
    */

    2. 解题思路


    3. 代码

     1 class Solution {
     2 public:
     3     bool isSameTree(TreeNode* p, TreeNode* q)
     4     {
     5         if (NULL == p && NULL == q)
     6         {
     7             return true;
     8         }
     9 
    10         if (NULL != p && NULL != q)
    11         {
    12             if (p->val == q->val)
    13             {
    14                 return isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
    15             }
    16             else
    17             {
    18                 return false;
    19             }
    20         }
    21         return false;
    22     }
    23 #pragma region 非递归方式
    24     bool isSameTree_1(TreeNode *p, TreeNode *q) 
    25     {
    26         if(!isSameNode(p, q))
    27             return false;
    28         if(!p && !q)
    29             return true;
    30 
    31         queue<TreeNode*> lqueue;
    32         queue<TreeNode*> rqueue;
    33         lqueue.push(p);
    34         rqueue.push(q);
    35         while(!lqueue.empty() && !rqueue.empty())
    36         {
    37             TreeNode* lfront = lqueue.front();
    38             TreeNode* rfront = rqueue.front();
    39 
    40             lqueue.pop();
    41             rqueue.pop();
    42 
    43             if(!isSameNode(lfront->left, rfront->left))
    44                 return false;
    45             if(lfront->left && rfront->left)
    46             {
    47                 lqueue.push(lfront->left);
    48                 rqueue.push(rfront->left);
    49             }
    50 
    51             if(!isSameNode(lfront->right, rfront->right))
    52                 return false;
    53             if(lfront->right && rfront->right)
    54             {
    55                 lqueue.push(lfront->right);
    56                 rqueue.push(rfront->right);
    57             }
    58         }
    59         return true;
    60     }
    61     bool isSameNode(TreeNode* p, TreeNode *q)
    62     {
    63         if(!p && !q)
    64             return true;
    65         if((p && !q) || (!p && q) || (p->val != q->val))
    66             return false;
    67         return true;
    68     }
    69 #pragma endregion
    70 };

    4. 反思

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  • 原文地址:https://www.cnblogs.com/whl2012/p/5596735.html
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