235. 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.

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题目：在平衡二叉树中寻找任意两个节点的公共祖先

思路：

利用while循环，就可以循环遍历BST树，

判断的条件如下：

如果p，q都小于root，那么在root的左子树中查找

如果p，q都大于root，那么在root的右子树中查找

如果p，q在root的两边，那么root就是要求得公共节点

如果q，q中有一个与root指针相同，那么返回root即可。

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code：

/**
 * 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) {
        TreeNode *tmp =root;
        while(tmp!=nullptr){
            if(p->val<tmp->val && q->val<tmp->val){
                tmp = tmp->left;
            }else if(p->val>tmp->val && q->val>tmp->val){
                tmp = tmp->right;
            }else if((p->val>tmp->val && q->val<tmp->val) || (p->val<tmp->val && q->val>tmp->val)){
                return tmp;
            }else if(p->val==tmp->val){
                return p;
            }else if(q->val==tmp->val){
                return q;
            }
        }//while
     return nullptr;   
    }//end-function
};