leetcode 101. Symmetric Tree

最新推荐文章于 2025-05-22 20:53:17 发布

anyiguan3906

最新推荐文章于 2025-05-22 20:53:17 发布

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文章标签： c/c++ python

原文链接：http://www.cnblogs.com/strongYaYa/p/6768138.html

版权

本文提供了一种使用递归方法和迭代方法来判断一棵二叉树是否为对称二叉树的有效算法。递归方法通过比较左右子树是否相等实现；迭代方法则利用栈保存节点对进行逐层对比。

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//递归的解法

/**
 * 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 == NULL)
            return true;
        TreeNode *pleft = root->left;
        TreeNode *pright = root->right;
        return isequal(pleft,pright);    
    }
    
    bool isequal(TreeNode *pleft, TreeNode *pright){
        if(pleft == NULL && pright == NULL )
            return true;
        if(pleft == NULL && pright != NULL)
            return false;
        if(pleft != NULL && pright == NULL)
            return false;
        if(pleft->val == pright->val)
            return isequal(pleft->left,pright->right) && isequal(pleft->right,pright->left);
        else{
            return false;
        }
    }
};

　　//循环的方法很悲哀不太会用C++ STL模板中的stack 和pair 所以用了python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root is None:
            return True
        
        stack = [[root.left,root.right]]
        
        while len(stack) > 0:
            pair = stack.pop()
            pleft = pair[0]
            pright = pair[1]
            
            if pleft is None and pright is None:
                continue
            if pleft is None or pright is None:
                return False
            if pleft.val == pright.val:
                stack.append([pleft.left,pright.right])
                stack.append([pleft.right,pright.left])
            else:
                return False
    
        return True