LeetCode | Same Tree

来源：互联网发布：linux中man的用法编辑：程序博客网时间：2024/06/10 17:08

题目：

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.

思路：

树的问题大部分都是递归求解。

代码：

递归实现：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool isSameTree(TreeNode *p, TreeNode *q) {        if(p == NULL && q == NULL){            return true;        }        else if(p != NULL && q != NULL && p->val == q->val){            return isSameTree(p->left, q->left) && isSameTree(p->right, q->right);        }        else{            return false;        }    }};

非递归实现：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool isSameTree(TreeNode *p, TreeNode *q) {        queue<TreeNode*> node1;        queue<TreeNode*> node2;                if(p == NULL && q == NULL){            return true;        }        else if(p != NULL && q != NULL && p->val == q->val){            node1.push(p);            node2.push(q);            while(!node1.empty() && !node2.empty()){                TreeNode* n1 = node1.front();                node1.pop();                TreeNode* n2 = node2.front();                node2.pop();                                if(n1->left != NULL && n2->left != NULL && n1->left->val == n2->left->val){                    node1.push(n1->left);                    node2.push(n2->left);                }                else if(!(n1->left == NULL && n2->left == NULL))                {                    return false;                }                if(n1->right != NULL && n2->right != NULL && n1->right->val == n2->right->val){                    node1.push(n1->right);                    node2.push(n2->right);                }                else if(!(n1->right == NULL && n2->right == NULL))                {                    return false;                }            }        }        else{            return false;        }        return true;    }};