Symmetric Tree[LeetCode]对称二叉树

https://leetcode.com/problems/symmetric-tree/

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.

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

   1
  /  2   3
    /
   4
         5

#include<iostream>
#include<queue>
#include<vector>
using namespace std;
struct TreeNode {
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

//层序遍历方法解决
class Solution {
public:
	bool isSymmetric(TreeNode *root) {
		queue<TreeNode *> iqueue; 
		vector<TreeNode *> ivec; //回文判断是否对称
		TreeNode teminateNode(INT_MAX); //使用INT_MAX代替#
		if(root!=NULL)
			iqueue.push(root);
		else
			return true; //空树是对称的，直接返回true
		while(!iqueue.empty()){ //层序遍历结束条件，其实是用不到的,可以改成while(true)
			while(!iqueue.empty()){ //遍历队列，直到为空
				TreeNode *temp=iqueue.front();
				iqueue.pop();
				if(temp->val!=INT_MAX){
<span style="white-space:pre">					</span>if(temp->left!=NULL)
<span style="white-space:pre">						</span>ivec.push_back(temp->left);
<span style="white-space:pre">					</span>else
<span style="white-space:pre">					</span>    ivec.push_back(&teminateNode);
<span style="white-space:pre">					</span>if(temp->right!=NULL)
<span style="white-space:pre">						</span>ivec.push_back(temp->right);
<span style="white-space:pre">					</span>else
<span style="white-space:pre">					</span>    ivec.push_back(&teminateNode);
<span style="white-space:pre">				</span>}

			}
			if(ivec.empty())  //如果树对称，则ivec肯定为空
				return true;
			//如果ivec不为空时，判断是否对称，实质判断是否为回文
			vector<TreeNode*>::iterator biter=ivec.begin();
			vector<TreeNode*>::iterator eiter=ivec.end()-1;
			while(biter<eiter){   
				if((*biter)->val!=(*eiter)->val)
					return false;//不对称，返回false
				biter++;
				eiter--;
			}
			//把ivec内容顺序压入队列，然后清空ivec
			for(int i=0;i<ivec.size();i++)
				iqueue.push(ivec[i]);
			ivec.clear();
		}
	}
};

递归解决方法：

class Solution {
public:
	bool isSymmetric(TreeNode *root) {
		if(root==NULL) //空树返回true
			return true;
		else 
			return isSymmetric(root->left,root->right);
	}
	bool isSymmetric(TreeNode* lroot,TreeNode* rroot){
		if(lroot==NULL && rroot==NULL) //全为空
			return true;
		if(lroot==NULL || rroot==NULL)//有一个为空，肯定不对称
			return false;
		if(lroot->val!=rroot->val)//值不相等肯定不对称
			return false;
		bool isLeft=isSymmetric(lroot->left,rroot->right);//判断左左和右右是否相等
		bool isRight=isSymmetric(lroot->right,rroot->left);//判断左右和右左是否相等
		return isLeft&&isRight; //都相等时，才为true
	}
};

Symmetric Tree[LeetCode]对称二叉树

原文：http://blog.csdn.net/sxhlovehmm/article/details/44676873