[leetcode-101]Symmetric Tree(c++）

最新推荐文章于 2022-08-13 11:52:24 发布

zdavb

最新推荐文章于 2022-08-13 11:52:24 发布

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本文介绍如何通过递归和迭代两种方式检查给定的二叉树是否为对称二叉树，即左右子树镜像对称。

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

For example, this binary tree is symmetric:

/ \
2 2
/ \ / \
3 4 4 3
But the following is not:
1
/ \
2 2
\ \
3 3
Note:
Bonus points if you could solve it both recursively and iteratively.

confused what “{1,#,2,3}” means? > read more on how binary tree is serialized on OJ.

代码如下: 递归版/4ms

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
bool checkSym(struct TreeNode* leftNode,struct TreeNode* rightNode){
    if(!leftNode || !rightNode)
        return leftNode==rightNode;
    if(leftNode->val != rightNode->val)
        return false;
    return checkSym(leftNode->right,rightNode->left) && checkSym(leftNode->left,rightNode->right);

}
bool isSymmetric(struct TreeNode* root) {
    if(!root)
        return true;
    return checkSym(root->left,root->right);
}

代码如下：迭代版/4ms
(维护一个队列或者栈)

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        queue<TreeNode*> queue;
        if(!root)
            return true;
        queue.push(root->left);
        queue.push(root->right);
        TreeNode *front,*second;

        while(!queue.empty()){
            front = queue.front();queue.pop();
            second = queue.front();queue.pop();
            if(!front || !second){
                if(front!=second)
                    return false;
                continue;
            }

            if(front->val!=second->val)
                return false;
            queue.push(front->right);
            queue.push(second->left);
            queue.push(front->left);
            queue.push(second->right);
        }
        return true;
    }
};