LeetCode OJ 之 Symmetric Tree（对称树）

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题目：

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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1  / \ 2   3    /   4    \     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

1、迭代版

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; *///迭代版，时间复杂度O(n)        bool isSymmetric(TreeNode *root)         {            if (!root) //空树，返回真                return true;            stack<TreeNode*> s;//使用栈保存树            s.push(root->left);            s.push(root->right);            while (!s.empty ()) //栈非空时            {                auto p = s.top (); //取出栈顶                s.pop();                auto q = s.top (); //取出栈顶                s.pop();                if (!p && !q) //p和q都为空时，结束此次循环，继续下次循环                    continue;                if (!p || !q) //p和q只有一个为空时，则不对称，结束                    return false;                if (p->val != q->val) //p和q都非空，但值不同时，不对称，结束                    return false;                s.push(p->left);//入栈顺序很重要                s.push(q->right);                s.push(p->right);                s.push(q->left);            }            return true;        }

2、递归版

//递归版，时间复杂度O(n)        bool isSymmetric(TreeNode *root)         {            //树空则返回真，否则执行下面的递归函数判断            return root ? isSymmetric(root->left, root->right) : true;        }        bool isSymmetric(TreeNode *left, TreeNode *right)         {            if (!left && !right)                 return true; // 递归结束条件，左结点和右结点都为空，返回真            if (!left || !right)                 return false; //递归结束条件，左结点和右结点只有一个为空，返回假            //递归，左右结点的值相等，且左结点的左和右结点的右，左结点的右和右结点的左继续递归            return left->val == right->val && isSymmetric(left->left, right->right)&& isSymmetric(left->right, right->left);        }

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