面試題 04.07. 首個共同祖先

設計並實現一個算法，找出二叉樹中某兩個節點的第一個共同祖先。不得將其他的節點存儲在另外的數據結構中。注意：這不一定是二叉搜索樹。

/**
 * 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 cover(TreeNode* root,TreeNode* t)
    {
        if(root == NULL)
            return NULL;
        if(root == t)
            return root;
        return cover(root->left,t) || cover(root->right,t);
    }
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
            if(!cover(root,p) || !cover(root,q))
                return NULL;
            if(root == NULL)
                return NULL;
            if(root == q || root == p)
                return root;
            bool is_leftq = cover(root->left,p);
            bool is_leftp = cover(root->left,q);

            if(is_leftq != is_leftp)
                return root;
            TreeNode* child_side = is_leftq ? root->left : root->right;
            return lowestCommonAncestor(child_side,p,q);
    }
};

 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
            if(root == NULL)
                return NULL;
            if(root == q && root == p)
                return root;
            TreeNode* x = lowestCommonAncestor(root->left,p,q);
            if(x != NULL && x != p && x != q)
            {
                return x;
            }
            TreeNode* y = lowestCommonAncestor(root->right,p,q);
            if(y != NULL && y != p && y != q)
            {
                return y;
            }
            if(x != NULL && y != NULL)
                return root;
            else if(root == p || root == q)
                return root;
            else 
                return x == NULL ? y : x;
    }