LeetCode 101. Symmetric Tree

原题链接在这里：https://leetcode.com/problems/symmetric-tree/

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:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following is not:

    1
   / \
  2   2
   \   \
   3    3

题解：

本题与Same Tree类似。这里比较是否symmetric, 也是用recursion, 需要写一个helper function, 递归调用，每次对称着比较.

Time Complexity: O(n), n是tree的node数目. Space: O(logn).

AC Java:

/**
 * 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){
            return true;
        }
        return isSymm(root.left, root.right);
    }
    
    private boolean isSymm(TreeNode p, TreeNode q){
        if(p == null && q == null){
            return true;
        }
        if(p == null || q == null){
            return false;
        }
        return p.val == q.val && isSymm(p.left, q.right) && isSymm(p.right, q.left);
    }
}

 AC 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 isSymmetric(TreeNode* root) {
        if(!root){
            return true;
        }

        return isSym(root->left, root->right);
    }

    bool isSym(TreeNode* p, TreeNode* q){
        if(!p && !q){
            return true;
        }

        if(!p || !q){
            return false;
        }

        return p->val == q->val && isSym(p->left, q->right) && isSym(p->right, q->left);
    }
};