leetcode：Lowest Common Ancestor of a Binary Search Tree

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.

分析：由于在所谓的二叉搜索树（binary search tree）中，处处满足顺序性（即任一节点的左（右）子树种，所有节点均小于（大于）r）。

思路： 1、当头结点为空时，返回空指针

2、如果节点p、q的值都比root的值要小，那么LCA一定为root的左子树；而如果节点p、q的值都比root的值要大，那么LCA一定为root的右子树；当节点p、q的值中一个比root的值要大，另一个比它小时，LCA就是root了。

3、我们要考虑是否能覆盖到节点是它自身子节点的情况，这时返回的是p或q的其中一个。

代码如下：（recursive solution）

/**
 * 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) {
        while(root !=nullptr){
        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;     // 当p->val = root->val或 q->val =root->val时，就是节点是它自身子节点的情况了        }      return root;    } };

　也可以简洁点：

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

其他参考解法：

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

        if( p->val < root->val && q->val < root->val)
            return lowestCommonAncestor(root->left, p, q);
        if( p->val > root->val && q->val > root->val)
            return lowestCommonAncestor(root->right, p, q);
    }
};

或：（iterative solution）　　

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