刷题101. Symmetric Tree

一、题目说明

题目101. Symmetric Tree，给定一个二叉树，判断是否对称。题目难度是Easy！

二、我的解答

这个题目不难，但是做错了2次（开始以为“前序遍历”和“后序遍历”对称即可，实际上不是）。递归方法的代码：

class Solution{
	public:
		bool isMirror(TreeNode* t1,TreeNode*t2){
			if(t1==NULL && t2==NULL) return true;
			if(t1==NULL || t2==NULL) return false;
			return (t1->val==t2->val) && isMirror(t1->left,t2->right) && isMirror(t1->right,t2->left);
		}
		bool isSymmetric(TreeNode* root){
			return isMirror(root,root);
		}
};

性能：

Runtime: 8 ms, faster than 52.13% of C++ online submissions for Symmetric Tree.
Memory Usage: 14.7 MB, less than 88.14% of C++ online submissions for Symmetric Tree.

三、优化措施

非递归算法，需要用一个队列queue，先判断根是否对称，然后判断左子树和右子树是否对称。

class Solution{
	public:
		//iterative
		bool isSymmetric(TreeNode* root){
			if(root==NULL || (root->left==NULL && root->right==NULL)){
				return true;
			}
			queue<TreeNode*> q;
			//将左右子树入队列
			q.push(root->left);
			q.push(root->right);
			while(!q.empty()){
			    //左子树出队列
				TreeNode* l = q.front();
				q.pop();
				//左子树出队列
				TreeNode* r = q.front();
				q.pop();
				if(l ==NULL && r==NULL){
					continue;
				}
				if(l==NULL || r==NULL){
					return false;
				}
				if(l->val != r->val){
					return false;
				}
				
				//将左子树的“左子树”，右子树的“右子树”入队列
				q.push(l->left);
				q.push(r->right);
				
				//将左子树的“右子树”，右子树的“左子树”入队列
				q.push(l->right);
				q.push(r->left);
			}
			
			return true;
		}
};

性能如下：

Runtime: 4 ms, faster than 85.48% of C++ online submissions for Symmetric Tree.
Memory Usage: 14.7 MB, less than 91.53% of C++ online submissions for Symmetric Tree.