LeetCode 题解(223) : Symmetric Tree

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本文详细介绍了使用递归和迭代方法解决判断二叉树是否为镜像的问题，提供了C++、Java和Python三种语言的代码实现。

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题目：

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

For example, this binary tree is symmetric:

But the following is not:

Note:
Bonus points if you could solve it both recursively and iteratively.

题解：

循环写的比较难看。

C++版：

/**
 * 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 || (root->left == NULL && root->right == NULL))
            return true;
        return isSymmetric(root->left, root->right);
    }
    
    bool isSymmetric(TreeNode* first, TreeNode* second) {
        if((first == NULL && second != NULL) || (first != NULL && second == NULL))
            return false;
        if(first == NULL && second == NULL)
            return true;
        if(first->val == second->val) {
            return isSymmetric(first->left, second->right) && isSymmetric(first->right, second->left);
        }
        return false;
    }
};

Java版：

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

Python版：

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

class Solution(object):
    def isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root == None or (root.left == None and root.right == None):
            return True
        if root.left != None and root.right != None and root.left.val == root.right.val:
            ql1, ql2, qr1, qr2 = [root.left], [], [root.right], []
            while len(ql1) != 0 or len(qr1) != 0:
                for i in range(len(ql1)):
                    if ql1[i].left != None and qr1[-1 - i].right != None:
                        if ql1[i].left.val != qr1[-1 - i].right.val:
                            return False
                    if (ql1[i].left != None and qr1[-1 - i].right == None) or (ql1[i].left == None and qr1[-1 - i].right != None):
                        return False
                    if ql1[i].right != None and qr1[-1 - i].left != None:
                        if ql1[i].right.val != qr1[-1 - i].left.val:
                            return False
                    if (ql1[i].right != None and qr1[-1 - i].left == None) or (ql1[i].right == None and qr1[-1 - i].left != None):
                        return False
                for i in ql1:
                    if i.left != None:
                        ql2.append(i.left)
                    if i.right != None:
                        ql2.append(i.right)
                        
                for i in qr1:
                    if i.left != None:
                        qr2.append(i.left)
                    if i.right != None:
                        qr2.append(i.right)
                ql1 = copy.copy(ql2)
                qr1 = copy.copy(qr2)
                ql2 = []
                qr2 = []
            return True
        else:
            return False