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.
Tips：找到一棵二叉查找树中两个结点的最低公共祖先。
解决方案：二叉查找树是一个已经排好序的二叉树，所有左子树结点都小于父结点，所有右子树结点都大于父节点，所以，只需要比较当前结点与p、q结点val的大小进行比较。若p、q的值都小于当前节点，那么最低公共祖先一定在当前节点的左子树上，相反则在右子树上。
方案一：使用循环：
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { while (true) { if (root.val > p.val && root.val > q.val) root = root.left; else if (root.val < p.val && root.val < q.val) root = root.right; else return root; } }
方案二： 使用递归：
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { TreeNode ans = root; if (root.val > p.val && root.val > q.val) { ans = lowestCommonAncestor(root.left, p, q); } else if (root.val < p.val && root.val < q.val) { ans = lowestCommonAncestor(root.right, p, q); } else { ans = root; } return ans; }
