[Leetcode][Tree][Symmetric Tree]

如果按层次遍历，存下每一层的点，会MLE。

1、递归版本：关键还是子问题的划分。

/**
 * 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 symmetric(TreeNode * leftnode, TreeNode * rightnode) {
        if (leftnode == NULL && rightnode == NULL) {
            return true;
        }
        if (leftnode == NULL || rightnode == NULL) {
            return false;
        }
        if (leftnode->val != rightnode->val) {
            return false;
        }
        return symmetric(leftnode->left, rightnode->right) &&
               symmetric(leftnode->right, rightnode->left);
    }
    bool isSymmetric(TreeNode *root) {
        if (root == NULL) {
            return true;
        } else {
            return symmetric(root->left, root->right);
        }
    }
};

2、迭代版本：思路很巧妙，不看题解自己真的想不出来。。对自己的智商好捉急。。

/**
 * 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) {
        queue<TreeNode *> q;
        if (root == NULL) {
            return true;
        } else {
            q.push(root->left);
            q.push(root->right);
        }
        while (!q.empty()) {
            TreeNode * n1 = q.front();
            q.pop();
            TreeNode * n2 = q.front();
            q.pop();
            if (n1 == NULL && n2 == NULL) {
                continue;
            }
            if (n1 == NULL || n2 == NULL) {
                return false;
            }
            if (n1->val != n2->val) {
                return false;
            }
            q.push(n1->left);
            q.push(n2->right);
            q.push(n1->right);
            q.push(n2->left);
        }
        return true;
    }
};