Lowest Common Ancestor of a Binary Search Tree（Java 递归与非递归）

题目描述：

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.

解法一：

/**方法一
*
* 给定一个二叉查找树（BST），寻找p和q的最小公共祖先结点
* 解题思路：首先找到到达p的路径和到达q的路径，将这两个路径存入队列中
* 然后对两个路径依次出队列，找出最后一组相同的结点，结束
* @param root
* @return
*/
public static TreeNode lowestCommonAncestor(TreeNode root,TreeNode p,TreeNode q){
if(root == null)
return null;
Queue<TreeNode> pPath = new LinkedList<TreeNode>();
Queue<TreeNode> qPath = new LinkedList<TreeNode>();
TreeNode pre = null;
//返回true，说明该结点存在；否则，说明给定的结点不存在，则无需再比较
if(findNode(root, p, pPath) && findNode(root, q, qPath)){
while(!pPath.isEmpty() && !qPath.isEmpty()){
TreeNode pNode = pPath.poll();
TreeNode qNode = qPath.poll();
if(pNode == qNode){ //记录最后一组相同结点
pre = pNode;
} else { //一旦不同，跳出循环
break;
}
}
}
return pre;
}

/**
* 在二叉树中查找结点p
* 将其祖先结点依次存入队列
* 返回true，说明该结点存在；否则，说明给定的结点不存在，则无需再比较
* @param root
* @param p
* @param stack
*/
public static boolean findNode(TreeNode root,TreeNode p,Queue<TreeNode> queue){
if(root == null)
return false;
if(p.val == root.val){
queue.add(root);
return true;
}
else if(Integer.valueOf(p.val.toString()) < Integer.valueOf(root.val.toString())){ //在左子树查找
queue.add(root);
return findNode(root.left, p, queue);
}
else {//在右子树查找
queue.add(root);
return findNode(root.right, p, queue);
}
}

解法二：

/** 方法二
*
* 递归算法
* @param root
* @param p
* @param q
* @return
*/
public static TreeNode LCA(TreeNode root,TreeNode p,TreeNode q){
if (root == null || p == null || q == null) return null;
if (Integer.valueOf(p.val.toString()) < Integer.valueOf(root.val.toString()) && Integer.valueOf(q.val.toString()) < Integer.valueOf(root.val.toString())) //两个结点都在左子树
return LCA(root.left, p, q);
else if (Integer.valueOf(p.val.toString()) > Integer.valueOf(root.val.toString()) && Integer.valueOf(q.val.toString()) > Integer.valueOf(root.val.toString())) //两个结点都在右子树
return LCA(root.right, p, q);
else
return root; //若两个结点一个在左子树上，另一个在右子树上，则共同祖先结点为当前的root
}