[Binary search tree] leetcode235: recent common ancestor binary search hormone (easy)

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Title:

Solution:

• 1) from the root node traversal binary search tree
• 2) If the right child node in the tree pq, places the right child node as a root Repeat step 1
• 3) If the node pq are left subtree, places left child node as a root Repeat step 1
• 4) If the node is not the same pq in a sub-tree, then there will be three cases of the following figure: pq the left and right sub-tree, the current root is the most recent common ancestor; pq in a node as a root, another sub-node, then p or q is at root node is a common ancestor of the
  
  code is as follows:

class Solution {
public:
    //解法1：递归版
    TreeNode* lowestCommonAncestor_1(TreeNode* root, TreeNode* p, TreeNode* q) {
        int parentVal=root->val;//根节点的值
        int pVal=p->val,qVal=q->val;//保存节点pq的值
        
        //若pq都在右子树中，则继续在右子树中寻找最近公共祖先
        if(pVal>parentVal&&qVal>parentVal)return lowestCommonAncestor(root->right,p,q);
        
        //若pq都在左子树中，则继续在左子树中寻找最近公共祖先
        else if(pVal<parentVal&&qVal<parentVal)return lowestCommonAncestor(root->left,p,q);
        
        //若pq位于左右子树中，那么根节点就是它们的最近公共祖先，或者说pq中的一个就为最近共同祖先
        else return root;
    }
    
    //解法二：迭代版
    TreeNode* lowestCommonAncestor_2(TreeNode* root,TreeNode* p,TreeNode* q)
    {
        int pVal=p->val,qVal=q->val;//保存节点pq的值
        TreeNode *node=root;//用指针node来遍历root为根节点的二叉搜索树
        while(node!=nullptr)
        {
            int parentVal=node->val;//根节点的值
            if(pVal>parentVal&&qVal>parentVal)node=node->right;//右子树中继续寻找
            else if(pVal<parentVal&&qVal<parentVal)node=node->left;//左子树中继续寻找
            else return node;
        }
        return nullptr;
    }
};