101. Symmetric Tree (C语言)

判断是否为左右镜面对称的二叉树

题目

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

Example 1:

Input: root = [1,2,2,3,4,4,3]
Output: true

Example 2:

Input: root = [1,2,2,null,3,null,3]
Output: false

Constraints:
The number of nodes in the tree is in the range [1, 1000].
-100 <= Node.val <= 100

解答

运用的是递归求解

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     struct TreeNode *left;*     struct TreeNode *right;* };*/bool isContrary(struct TreeNode* p, struct TreeNode* q)
{if(p == NULL && q == NULL)return true;else if((p == NULL && q != NULL) || (p != NULL && q == NULL))return false;elseif(p->val != q->val)return false;return (isContrary(p->left, q->right) && isContrary(p->right, q->left));
}bool isSymmetric(struct TreeNode* root){if(root == NULL) return true;return isContrary(root->left, root->right);
}

总结

运用了和“判定两个二叉树是否相同”类似的思路，一定要把握好哪些特征是相同的。在这道题中转化为，判定根节点的左子树和右子树是否是完全相反的。
“判定两个二叉树是否相同”：link

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101. Symmetric Tree (C语言)

101. Symmetric Tree (C语言)

题目

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总结

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