博文中会简要介绍Leetcode P0101题目分析及解题思路。
“Symmetric Tree”是一道比较基础的树问题，深度优先搜索即可解决。
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 3But the following [1,2,2,null,3,null,3] is not:
1 / \ 2 2 \ \ 3 3
以下是Java的题解代码实现。
/**
* 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 isSymmetric(TreeNode root) {
if (root == null)
return true;
return this.validate(root.left, root.right);
}
private boolean validate(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 this.validate(left.left, right.right) && this.validate(left.right, right.left);
}
}
以下是C++的题解代码实现。
/**
* 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 isSymmetric(TreeNode* root) {
if (root == NULL)
return true;
return this->Validate(root->left, root->right);
}
private:
bool Validate(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 this->Validate(left->left, right->right) && this->Validate(left->right, right->left);
}
};