【LeetCode】Symmetric Tree

description:

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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

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}"

code : (build a symmetric tree first, then check if the two trees are the same )

/** * 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) {        // Note: The Solution object is instantiated only once and is reused by each test case.        if(root == NULL)             return true;        TreeNode *syroot = new TreeNode(root->val);        build(syroot->left,root->right);        build(syroot->right,root->left);        if(isSameTree(syroot,root))            return true;        else return false;            }    void build(TreeNode *&syroot,TreeNode *root)    {        if(root == NULL)        {            syroot = NULL;            return;        }        syroot = new TreeNode(root->val);        build(syroot->left,root->right);        build(syroot->right,root->left);            }    bool isSameTree(TreeNode *p, TreeNode *q)     {          // Start typing your C/C++ solution below          // DO NOT write int main() function          if(p == NULL && q == NULL) return true;          if((p == NULL && q !=NULL) || (p !=NULL && q == NULL)) return false;          if(p->val != q->val) return false;          if(!issame(p->left,q->left)) return false;          if(!issame(p->right,q->right)) return false;          return true;                }      bool issame(TreeNode *p, TreeNode *q)      {          if((p == NULL && q !=NULL) || (p !=NULL && q == NULL)) return false;          if(p == NULL && q == NULL) return true;          if(p->val != q->val) return false;          if(!issame(p->left,q->left)) return false;          if(!issame(p->right,q->right)) return false;          return true;                } };

下面的代码更简洁，不用构造一颗对称树：但是时间复杂度并没有优化多少

/** * 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) {        // Note: The Solution object is instantiated only once and is reused by each test case.        if(root == NULL)             return true;        return isSame(root,root);            }    bool isSame(TreeNode *syroot,TreeNode *root)    {        if(syroot == NULL && root == NULL)            return true;        else if(syroot == NULL || root == NULL)            return false;        if(syroot->val != root->val)            return false;        return isSame(syroot->left,root->right) && isSame(syroot->right,root->left);    }     };