HappyLeetcode43:Symmetric Tree

標籤：

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

    1   /   2   2 / \ / 3  4 4  3

But the following is not:

    1   /   2   2   \      3    3

這道題目可以用迭代的方式做，也可以用遞迴的方式做。遇見和樹相關的題目，第一反應是遞迴，所以我的解法是遞迴。

代碼奉上：

class Solution { public:     bool IsSymmetric(TreeNode *node1, TreeNode *node2)     {         if (node1 == NULL&&node2 == NULL)             return true;         if (node1 == NULL||node2 == NULL)             return false;         bool result1, result2;         if (node1->val != node2->val)             return false;         else         {             result1 = IsSymmetric(node1->left, node2->right);             result2 = IsSymmetric(node1->right, node2->left);         }         return result1&&result2;     }     bool isSymmetric(TreeNode *root) {         if (root == NULL || root->left == NULL&&root->right == NULL)             return true;         bool result = IsSymmetric(root->left, root->right);         return result;     } };

看了相關人的解法，有的人直接用迭代的方式解出來了，感覺很受啟發

/** * 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 isSymmetric(TreeNode *root) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        if(root == NULL)return true;        queue<TreeNode*> qleft, qright;        if(root->left)qleft.push(root->left);        if(root->right)qright.push(root->right);        while(qleft.empty() == false && qright.empty() == false)        {            TreeNode *ql = qleft.front();            TreeNode *qr = qright.front();            qleft.pop();  qright.pop();            if(ql->val == qr->val)            {                if(ql->left && qr->right)                {                    qleft.push(ql->left);                    qright.push(qr->right);                }                else if(ql->left || qr->right)                    return false;                if(qr->left && ql->right)                {                    qleft.push(qr->left);                    qright.push(ql->right);                }                else if(qr->left || ql->right)                    return false;            }            else return false;        }        if(qleft.empty() && qright.empty())            return true;        else return false;    }};

轉自http://www.cnblogs.com/TenosDoIt/p/3440729.html

HappyLeetcode43:Symmetric Tree