235. Lowest Common Ancestor of a Binary Search Tree(LCA最低公共祖先)

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 2and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

思路：因为是bst, 如果root，的值介于pq之间，root就是共同祖先。

root不是共同祖先，则递归地判断左右子树是不是公共祖先。

如果两个节点值都小于根节点，说明他们都在根节点的左子树上，我们往左子树上找
如果两个节点值都大于根节点，说明他们都在根节点的右子树上，我们往右子树上找
如果一个节点值大于根节点，一个节点值小于根节点，说明他们他们一个在根节点的左子树上一个在根节点的右子树上，那么根节点就是他们的最近公共祖先节点。
画个图看一下，比如要找0和5的最近公共祖先节点，

235. Lowest Common Ancestor of a Binary Search Tree(LCA最低公共祖先)

递归版：

1 class Solution {
2 public:
3     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
4         if(root->val > p->val && root->val > q->val) return lowestCommonAncestor(root->left,p,q);
5         if(root->val < p->val && root->val < q->val) return lowestCommonAncestor(root->right,p,q);
6         return root;   
7     }
8 };

循环版：

1 class Solution {
2     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
3         while((root.val-p.val)*(root.val-q.val)>0)
4             root = ((root.val-p.val)>0?root.left:root.right);
5         return root;
6     
7     }
8 }