【LeetCode】三道简单的递归问题

1.single-number

Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

class Solution {
public:
    int singleNumber(int A[], int n) {
        int rs = 0;
        
        for (int i = 0; i < n; i++)
        {
            rs ^= A[i];
        }
        return rs;
    }
};

2.maximum-depth-of-binary-tree

Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode *root) {
        
        if (NULL == root)
        	return 0;
        int l = maxDepth(root->left);
        int r = maxDepth(root->right);
        return l > r ? l + 1 : r + 1;
    }
};

3.same-tree

Given two binary trees, write a function to check if they are equal or not.

Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSameTree(TreeNode *p, TreeNode *q) {
        
        if (NULL == p && NULL == q)
            return true;
        if (NULL == p || NULL == q)
            return false;
        if (p->val == q->val)
        {
            bool rs1, rs2;
            rs1 = isSameTree(p->left, q->left);
            rs2 = isSameTree(p->right, q->right);
            if (rs1 && rs2)
            	return true;
        }
        return false;    
    }
};