101. Symmetric Tree*

101. Symmetric Tree*

https://leetcode.com/problems/symmetric-tree/

题目描述

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.

C++ 实现 1

先用最简单的递归方法来解决, 后面介绍迭代的方法. 对于递归的方法, 核心在于判断两个子树是否对称相等.

/**
 * 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 {
private:
    bool isSymmetricSame(TreeNode *a, TreeNode *b) {
        if (!a || !b) return a == b;
        return (a->val == b->val) && 
            isSymmetricSame(a->right, b->left) &&
            isSymmetricSame(a->left, b->right);
    }
public:
    bool isSymmetric(TreeNode* root) {
        if (!root) return true;
        return isSymmetricSame(root->left, root->right);
    }
};

C++ 实现 2

利用迭代的方法, 使用层序遍历; 使用两个队列, 一个队列保留左子树, 另一个队列保留右子树, 这个代码看起来比较复杂, C++ 实现 3 的逻辑更为简单, 因为她将空指针也放入到队列中了.

/**
 * 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) return true;
        if (!root->left || !root->right) return root->left == root->right;
        queue<TreeNode*> q1, q2;
        q1.push(root->left);
        q2.push(root->right);
        while (!q1.empty() || !q2.empty()) {
            auto size1 = q1.size();
            auto size2 = q2.size();
            // 每一层级节点个数要相等
            if (size1 != size2) return false;
            while (size1 --) {
                auto r1 = q1.front();
                q1.pop();
                auto r2 = q2.front();
                q2.pop();
                // 同层级对应的节点值不同的情况
                if (r1->val != r2->val) return false;
                if (r1->left) {
                    if (!r2->right || r2->right->val != r1->left->val) return false;
                    q1.push(r1->left);
                    q2.push(r2->right);
                } else if (r2->right) return false; // 左孩子不存在而右孩子存在的情况
                if (r1->right) {
                    if (!r2->left || r2->left->val != r1->right->val) return false;
                    q1.push(r1->right);
                    q2.push(r2->left);
                } else if (r2->left) return false;
            }
        }
        return true;
    }
};

C++ 实现 3

使用队列的方法. 注意, 空指针也是可以放入到队列中的. 根节点存入队列两次. 注意这个代码还可以进一步简化, root 无需存入队列两次; 如果存入两次, 相当于整棵树要被检查两遍. 在 C++ 实现 4 做了进一步简化.

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if (!root) return true;
        if (!root->left || !root->right) return root->left == root->right;
        queue<TreeNode*> q;
        q.push(root);
        q.push(root);
        while (!q.empty()) {
            auto a = q.front();
            q.pop();
            auto b = q.front();
            q.pop();
            if (!a && !b) continue;
            if (!a || !b) return false;
            if (a->val != b->val) return false;
            q.push(a->left);
            q.push(b->right);
            q.push(a->right);
            q.push(b->left);
        }
        return true;
    }
};

C++ 实现 4

/**
 * 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) return true;
        queue<TreeNode*> q;
        q.push(root->left);  // 直接存入左右孩子, 不管它们是不是空指针
        q.push(root->right);
        while (!q.empty()) {
            auto a = q.front();
            q.pop();
            auto b = q.front();
            q.pop();
            if (!a && !b) continue;
            if (!a || !b) return false;
            if (a->val != b->val) return false;
            q.push(a->left);
            q.push(b->right);
            q.push(a->right);
            q.push(b->left);
        }
        return true;
    }
};