Symmetric Tree

最新推荐文章于 2024-01-17 14:31:46 发布

东邪岛主止水

最新推荐文章于 2024-01-17 14:31:46 发布

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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，并详细解释了如何通过比较节点值来判断整棵树是否为镜像。

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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

递归形式：

/**
 * 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 symmetric(TreeNode* left,TreeNode* right){
        if(!left && !right)
            return true;
        if(left && right){
            if(left->val != right->val)
                return false;
            return symmetric(left->left,right->right)&&symmetric(left->right,right->left);
        }
        return false;
    }
public:
    bool isSymmetric(TreeNode* root) {
        if(!root)
            return true;
        return symmetric(root->left,root->right);

    }
};

迭代形式：

/**
 * 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;
        stack<TreeNode*> left,right;
        left.push(root->left);
        right.push(root->right);
        while(!left.empty()&&!right.empty()){
            TreeNode* lf = left.top();
            left.pop();
            TreeNode* rg = right.top();
            right.pop();
            if(!lf&&!rg)
                continue;
            if((lf&&!rg )||(!lf&&rg) )
                return false;
            if(lf->val!=rg->val)
                return false;
            left.push(lf->left);
            right.push(rg->right);
            left.push(lf->right);
            right.push(rg->left);

        }
        return true;

    }
};