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.

由于这是一颗二叉搜索树,我们根据nodes v and w值的大小,能确定他们位于根节点的左边还是右边。如果两个节点都在根节点的左子树上,那么他们的LCA也肯定在根节点的左子树;如果两个节点都在根节点的右子树上,那么他们的LCA也肯定在根节点的右子树;如果两个节点分别位于根结的左右子树,那么根节点就是他们的LCA。否则的话直接返回根节点即可。代码如下:

 /**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(root==null || p==null || q==null) return null; if(p.val<root.val && q.val<root.val){ //在根节点左边
return lowestCommonAncestor(root.left, p, q);
}
else if(p.val>root.val && q.val>root.val){ //在根节点右边
return lowestCommonAncestor(root.right, p, q);
}
else return root; //在根节点两侧,或者其中一个节点与根节点相等
}
}

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