二叉树系列 - [LeetCode] Symmetric Tree 判断二叉树是否对称，递归和非递归实现

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.

 

二叉树是否对称的本质，其实是判定两棵树是否镜像。

递归是很常见的实现方式，最简便。

/**

 * 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 compRoot(root -> left, root -> right);

    }

private:

    bool compRoot(TreeNode* lroot, TreeNode* rroot){

        if(!lroot) return (NULL == rroot);

        if(NULL == rroot) return false;

        if(lroot -> val != rroot -> val) return false;

        return (compRoot(lroot -> left, rroot -> right) && compRoot(lroot -> right, rroot -> left));

    }

};

 

非递归，我的方法其实还是很常规，用栈来代替。因为是对称比较，所以要两个栈。

这个思路其实可以稍微简化一下，改用一个双端队列deque实现。比起用两个栈来，显得稍微“洋气”一点 ==。

/**

 * 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(!root -> left && !root -> right) return true;

        if( (!root -> left && root -> right) || (root -> left && !root -> right) ) return false;

        deque<TreeNode*> dq;

        dq.push_front(root -> left);

        dq.push_back(root -> right);

        while(!dq.empty()){

            TreeNode* lroot = dq.front();

            TreeNode* rroot = dq.back();

            dq.pop_front();

            dq.pop_back();

            if(lroot -> val != rroot -> val) return false;

            if( (!lroot -> right && rroot -> left) || (lroot -> right && !rroot -> left) ) return false;

            if(lroot -> right){

                dq.push_front(lroot -> right);

                dq.push_back(rroot -> left);

            }

            if( (!lroot -> left && rroot -> right) || (lroot -> left && !rroot -> right) ) return false;

            if(lroot -> left){

                dq.push_front(lroot -> left);

                dq.push_back(rroot -> right);

            }

        }

        return true;

    }

};