235.236：Lowest Common Ancestor of a Binary Search Tree/Tree(LCA问题)

连续两道LCA问题，一道是BST树的LCA求解，另外一道是普通二叉树的LCA求解。

235：

Total Accepted: 119152
Total Submissions: 311598
Difficulty: Easy
Contributors: Admin
Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5
For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

该题直接秒杀，递归版本：

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

非递归版本：

/**
 * 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:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(p == NULL || q == NULL)
            return NULL;
        while(root){
            if(p->val < root->val && q->val < root->val)
                root = root->left;
            else if(p->val > root->val && q->val > root->val)
                root = root->right;
            else
                break;
        }
        return root;
    }
};

236：

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

同样采用递归，不过这次不能利用BST的特性了。我们分别在查找某个节点的所有节点看它的子节点是否有p或q即可。