package cn.edu.xidian.sselab;
/**
* title:Lowest Common Ancestor of a Binary Search Tree
* content:
* 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.
*/
public class LowestCommonAncestor {
public class TreeNode
{
int val;
TreeNode left;
TreeNode right;
TreeNode(int x)
{
val = x;
}
}
/**
* 二叉排序树又叫二叉查找树，英文名称是：Binary Sort Tree. Binary Search Tree
* BST的定义就不详细说了，我用一句话概括：左 < 中 < 右。 根据这个原理，我们可以推断：BST的中序遍历必定是严格递增的。
* 先序遍历；中序遍历；后续遍历；层次遍历。事实上，知道任意两种方式，并不能唯一地确定树的结构，
* 但是，只要知道中序遍历和另外任意一种遍历方式，就一定可以唯一地确定一棵树
*
* 本题目是求两个子节点的最小公共父节点，采用中序遍历的方式，因为是二叉排序树，所以无形中给了一个已知的条件
* 左子树小于父节点，小于右子树
* 如果如果p,q 比root小, 则LCA必定在左子树, 如果p,q比root大, 则LCA必定在右子树. 如果一大一小, 则root即为LCA.
*/
public TreeNode lowestCommonAncestor(TreeNode root,TreeNode p,TreeNode q){
TreeNode lowestCommonAncestor = root;
if(root == null){
return root;
}else if(p == null || q == null){
return root;
}else if((p.val < root.val && q.val > root.val)||(p.val > root.val && q.val < root.val)){
return root;
}else if(p.val < root.val && q.val < root.val){
lowestCommonAncestor = lowestCommonAncestor(root.left, p, q);
}else if(p.val > root.val && q.val > root.val){
lowestCommonAncestor = lowestCommonAncestor(root.right, p, q);
}
return lowestCommonAncestor;
}
}