[leedcode 235] Lowest Common Ancestor of a Binary Search Tree

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.

/**
 * 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) {
        //借助二叉搜索树的性质，左子树小于根节点，右子树大于根节点的性质，对比p，q的值跟root值的关系，可得到下面的做法
        if(root==null||p==null||q==null) return null;
        if(p==root||q==root) return root;
        int temp=root.val;
        if(temp>p.val&&temp>q.val){
            return lowestCommonAncestor(root.left,p,q);
        }else if(temp<p.val&&temp<q.val)
            return lowestCommonAncestor(root.right,p,q);
        return root;
    }
}