1.描述

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1
   / \
  2   2
   \   \
   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

2.方法一

2.1分析

將右子樹反轉，然后判斷左子樹和反轉后的右子樹是否相同。

2.2代碼

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

void reversal(struct TreeNode* root) {
    if (NULL == root) return ;
    reversal(root->left);
    reversal(root->right);
    struct TreeNode* tmp;
    tmp = root->left;
    root->left = root->right;
    root->right = tmp;
}

bool isSameTree(struct TreeNode* tree1, struct TreeNode* tree2) {
    if (NULL == tree1 && NULL == tree2) return true;
    if (NULL == tree1 || NULL == tree2) return false;
    if (tree1->val != tree2->val) return false;
    bool left = isSameTree(tree1->left, tree2->left);
    bool right = isSameTree(tree1->right, tree2->right);
    return left && right;
}
 
bool isSymmetric(struct TreeNode* root) {
    if (NULL == root) return true;
    if (NULL == root->left && NULL == root->right) return true;
    if (NULL == root->left || NULL == root->right) return false;
    reversal(root->right);
    return isSameTree(root->left, root->right);
}

3.方法二

3.1分析

層次遍歷二叉樹，判斷每層的元素是否對稱。

3.2代碼

4.方法三

4.1分析

簡單遞歸

4.2代碼

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
 
bool isSymmetricRL(struct TreeNode* root1, struct TreeNode* root2) {
    if (NULL == root1 && NULL == root2) return true;
    if (NULL == root1 || NULL == root2) return false;
    if (root1->val != root2->val) return false;
    return isSymmetricRL(root1->left, root2->right) && isSymmetricRL(root1->right, root2->left);
}

bool isSymmetric(struct TreeNode* root) {
    if (NULL == root) return true;
    return isSymmetricRL(root->left, root->right);
}