Leetcode:100. Same Tree

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Description

这里写图片描述

解题思路

采用DFS递归式的查找比较，若有一个不相等则返回false

/** * Definition for a binary tree node. * 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 (p==q);        return (p->val == q->val && isSameTree(p->left,q->left) && isSameTree(p->right,q->right));    }};

非递归解法，利用栈，将相同的数压入栈，若不相同或一方为空则返回false，到下一个数时pop到上一个数，若两棵树相等则遍历结束时必然两个站均为空。

/** * Definition for a binary tree node. * 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) {        stack<TreeNode*> stk1;        stack<TreeNode*> stk2;        stk1.push(p);        stk2.push(q);        while(!stk1.empty() && !stk2.empty()){            TreeNode* tmp1 = stk1.top();            TreeNode* tmp2 = stk2.top();            stk1.pop();            stk2.pop();            if(tmp1 && tmp2){                if(tmp1->val != tmp2->val)                    return false;                stk1.push(tmp1->left);                stk1.push(tmp1->right);                stk2.push(tmp2->left);                stk2.push(tmp2->right);            }            if((tmp1 && !tmp2) || (!tmp1 && tmp2))                return false;        }        return stk1.empty() == stk2.empty();    }};

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