Symmetric Tree

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

 

Note:

Bonus points if you could solve it both recursively and iteratively.

 

不管是递归还是非递归，找到关系就好做。

所谓对称，也就是：

1、left对应right

2、left->left对应right->right

3、left->right对应right->left

 

解法一：递归

/** * 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) {        if(!root)            return true;                    return isSymTree(root->left, root->right);    }    bool isSymTree(TreeNode* p, TreeNode* q)    {        if(!isSameNode(p, q))            return false;        if(!p && !q)            return true;        return isSymTree(p->left, q->right) && isSymTree(p->right, q->left);    }    bool isSameNode(TreeNode* p, TreeNode* q)    {        if(!p && !q)            return true;        if((!p && q) || (p && !q) || (p->val != q->val))            return false;        return true;    }};

 

解法二：非递归

使用两个队列，对左右子树分别进行层次遍历。

进队时的对应元素比较即可。

/** * 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) {        if(!root)            return true;                    if(!isSameNode(root->left, root->right))            return false;        if(!root->left && !root->right)            return true;                    queue
     
       lqueue;        queue
      
        rqueue;        lqueue.push(root->left);        rqueue.push(root->right);        while(!lqueue.empty() && !rqueue.empty())        {            TreeNode* lfront = lqueue.front();            TreeNode* rfront = rqueue.front();            lqueue.pop();            rqueue.pop();                        if(!isSameNode(lfront->left, rfront->right))                return false;            if(lfront->left && rfront->right)            {                lqueue.push(lfront->left);                rqueue.push(rfront->right);            }                        if(!isSameNode(lfront->right, rfront->left))                return false;            if(lfront->right && rfront->left)            {                lqueue.push(lfront->right);                rqueue.push(rfront->left);            }        }        return true;    }    bool isSameNode(TreeNode* p, TreeNode* q)    {        if(!p && !q)            return true;        if((!p && q) || (p && !q) || (p->val != q->val))            return false;        return true;    }};