面试笔试整理6：常见面试编程题

1、二叉树公共父节点
leecode236
递归解法：

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root==NULL || root==p||root==q) return root;
        TreeNode *left = lowestCommonAncestor(root->left, p, q);
        TreeNode *right = lowestCommonAncestor(root->right, p, q);
        if(left && right) return root;
        return left?left:right;
    }
};

非递归解法，最简单就是记录节点的父节点和其中一条路径：

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        stack = [root]
        pair = {root:None}
        while p not in pair or q not in pair:
            node = stack.pop()
            if node.left:
                stack.append(node.left)
                pair[node.left] = node
            if node.right:
                stack.append(node.right)
                pair[node.right] = node
        #找到到达p的路径
        path = set()
        while p :
            path.add(p)
            p = pair[p]
        while q not in path:
            q = pair[q]
        return q 

2、树的中序遍历

//非递归中序遍历
void inorderTraversal(TreeNode *root, vector<int> &path)
{
    stack s;
    TreeNode *p = root;
    while(p != NULL || !s.empty())
    {
        while(p != NULL)
        {
            s.push(p);
            p = p->left;
        }
        if(!s.empty())
        {
            p = s.top();
            path.push_back(p->val);
            s.pop();
            p = p->right;
        }
    }
}

递归版：

void inorder(TreeNode *root, vector<int> &path)
{
    if(root != NULL)
    {
        inorder(root->left, path);
        path.push_back(root->val);
        inorder(root->right, path);
    }
}

3、求两个已排序数组的交集
对于两个排序数组很简单，分别找两个指针指向数组头部，如果一个比另一个小就移动小的；如果相等就保存，同时两个指针都移动

public LinkedList intersection(int[] A, int[] B) {  
    if (A == null || B == null || A.length == 0 || B.length == 0) return null;  
    LinkedList list = new LinkedList();  
    int pointerA = 0;  
    int pointerB = 0;  
    while (pointerA < A.length && pointerB < B.length) {  
        if (A[pointerA] < B[pointerB]) pointerA++;  
        else if (A[pointerA] > B[pointerB]) pointerB++;  
        else {  
            list.add(A[pointerA]);  
            pointerA++;  
            pointerB++;  
        }  
    }  

    return list;  

}  

对于多个数组的情况同样可以用上面的方法或者：首先找到最短的数组，接着对这个数组中的值用二分查找在其他数组中寻找。