LeetCode刷题8--对称二叉树

101. 对称二叉树

链接

题目描述

给定一个二叉树，检查它是否是镜像对称的。

代码

python
递归解法

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root==None:
            return True
        if root.left==None and root.right==None:
            return True
        def helper(left,right):
            if left==None and right==None:
                return True
            if left==None or right==None:
                return False
            if left.val!=right.val:
                return False
            return helper(left.left,right.right) and helper(left.right,right.left)

        return helper(root.left,root.right)

非递归解法

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root==None:
            return True
        if root.left==None and root.right==None:
            return True
        if root.left==None or root.right==None:
            return False
        left=[root.left]
        right=[root.right]
        while(len(left) and len(right)):
            l=left.pop(0)
            r=right.pop(0)
            
            if l.val!=r.val:
                return False
            if l.left and r.right:
                left.append(l.left)
                right.append(r.right)
            elif l.left or r.right:
                    return False
            if l.right and r.left:
                
                left.append(l.right)
                right.append(r.left)
            elif l.right or r.left:
                    return False
           
        if len(left)==0 and len(right)==0:
            return True
        else:
            return False

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
     
public:
    bool helper(TreeNode* left,TreeNode* right){
     
        if (left==NULL && right==NULL) return true;
        if (left==NULL || right==NULL) return false;
        if (left->val!=right->val) return false;
        return helper(left->left,right->right) && helper(left->right,right->left);
    }
    bool isSymmetric(TreeNode* root) {
     
        if (root==NULL) return true;
        return helper(root->left,root->right);

    }
};

java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
     
    public boolean helper(TreeNode left,TreeNode right){
     
        if (left==null && right==null) return true;
        if (left==null || right==null) return false;
        if (left.val!=right.val) return false;
        return helper(left.left,right.right) && helper(left.right,right.left);
    }
    public boolean isSymmetric(TreeNode root) {
     
        if (root==null) return true;
        return helper(root.left,root.right);

    }
}