Leetcode 101. Symmetric Tree

最新推荐文章于 2025-09-11 19:19:55 发布

ruihuank

最新推荐文章于 2025-09-11 19:19:55 发布

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分类专栏： leetcode 笔记文章标签： leetcode

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本文介绍了一种检查二叉树是否为中心对称的方法，提供了递归和迭代两种实现方式。递归方法通过比较左子树和右子树来判断，而迭代方法则使用两个栈来跟踪节点。

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.

/**
 * Definition for a binary tree node.
 * 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 == nullptr) return true;
        return isSame(root->left, root->right);
    }

private:
    bool isSame(TreeNode* r1, TreeNode* r2) {
        if (r1 == nullptr || r2 == nullptr) return r2 == r1;
        return isSame(r1->left, r2->right) && r1->val == r2->val && isSame(r1->right, r2->left);
    }
};

/**
 * Definition for a binary tree node.
 * 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 == nullptr) return true;

        stack<TreeNode*> s1;
        stack<TreeNode*> s2;

        TreeNode* p1 = root->left;
        TreeNode* p2 = root->right;

        while (p1 != nullptr || !s1.empty()) {
            if (p1 == nullptr) {
                if (p2 != nullptr) return false;
                p1 = s1.top();
                s1.pop();
                p2 = s2.top();
                s2.pop();
                if (p1->val != p2->val) return false;
                p1 = p1->right;
                p2 = p2->left;
            }
            else {
                if (p2 == nullptr) return false;
                s1.push(p1);
                s2.push(p2);
                p1 = p1->left;
                p2 = p2->right;
            }
        }

        if (p2 == nullptr && s2.empty()) return true;
        return false;
    }
};