Leetcode - Tree - Symmetric Tree

最新推荐文章于 2022-06-04 21:05:23 发布

原创最新推荐文章于 2022-06-04 21:05:23 发布 · 510 阅读

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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，提供了两种实现方式：递归和迭代。递归方法通过比较左右子树是否相等来判断；迭代方法使用栈结构存储节点，逐步对比左右子树的对应节点。

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

/**
 * Definition for binary tree
 * 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==NULL)
            return true;
        return isEqual(root->left,root->right);
    }
    bool isEqual(TreeNode *l, TreeNode *r)
    {
        if(l==NULL&&r==NULL)
            return true;
        if(l==NULL&&r!=NULL||l!=NULL&&r==NULL)
            return false;
        if(l->val==r->val&&isEqual(l->left,r->right)&&isEqual(l->right,r->left))
                return true;
        else
            return false;
    }
};

/**
 * Definition for binary tree
 * 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==NULL)
            return true;
        stack<TreeNode*> nodeStack;
        nodeStack.push(root->left);
        nodeStack.push(root->right);
        TreeNode *lhs=NULL,*rhs=NULL;
        while(nodeStack.size()>1)
        {
            lhs=nodeStack.top();
            nodeStack.pop();
            rhs=nodeStack.top();
            nodeStack.pop();
            if(!lhs&&!rhs) continue;
            if((lhs&&!rhs)||(!lhs&&rhs)) return false;
            if(lhs->val!=rhs->val)return false;
            nodeStack.push(lhs->left);
            nodeStack.push(rhs->right);
            nodeStack.push(lhs->right);
            nodeStack.push(rhs->left);
        }
        return true;
    }
};