Symmetric Tree @leetcode

!本题分为迭代版和递归版，递归版比较直接，按照树遍历即可。跌代版这里创建了两个队列，每遇到一对节点（初始为根节点的左右子）：

1.先对比这两个节点的值是否相等：不相等则结束循环，相等则进入2

2.把它们的左右子节点，交叉安插到自己的队列中，然后对队列里的front，作为一对节点，走1

!iterative

if(!root) return true;
        queue<TreeNode*> queueStep;
        queue<TreeNode*> queueStep2;
        TreeNode *tn1,*tn2;
        tn1 = root->left;
        tn2 = root->right;
        while(1){
            if(!tn1 && tn2) return false;
            if(!tn2 && tn1) return false;
            if(tn1 && tn2 && tn1->val != tn2->val) return false;
            
            if(tn1){
                queueStep.push(tn1->left);
                queueStep.push(tn1->right);
            }
            if(tn2){
                queueStep2.push(tn2->right);
                queueStep2.push(tn2->left);
            }
            if(queueStep.size() == 0 && queueStep2.size() == 0) return true;
            /*else if(queueStep.size() == 0) return false;
            else if(queueStep2.size() == 0) return false;*/

            tn1 = queueStep.front();
            queueStep.pop();

            tn2 = queueStep2.front();
            queueStep2.pop();
        }

!recursive

/**
 * 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 == NULL) return true;
        return isSymmetricSmall(root->left,root->right);
    }

    bool isSymmetricSmall(TreeNode *left,TreeNode *right){
        if(left == NULL && right == NULL) return true;
        else if(left == NULL) return false;
        else if(right == NULL) return false;

        if(left->val != right->val) return false;
        
        if(isSymmetricSmall(left->right,right->left) && isSymmetricSmall(left->left,right->right))
            return true;
        else
            return false;
    }
};