235.二叉搜索数的最近公共祖先

方法：二次遍历

先遍历得到从根节点到p、根节点到q的路径，然后再遍历得到公共节点

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        List<TreeNode> path_p = getPath(root,p);
        List<TreeNode> path_q = getPath(root,q);
        TreeNode ancetor = null;
        for(int i = 0; i < path_p.size() && i < path_q.size();i++){
           if(path_p.get(i) ==  path_q.get(i)){
               ancetor = path_p.get(i);
           } else{
               break;
           }
        }   
        return ancetor;
    }

    //找到跟节点到目标节点的路径
    public List<TreeNode> getPath(TreeNode root, TreeNode target){
        List<TreeNode> path = new ArrayList<>();
        TreeNode node = root;
        while(node!=target){
            path.add(node);
            if(target.val < node.val){
                node = node.left;
            }else{
                node = node.right;
            }
        }
        path.add(node);
        return path;
    }
}

方法二：递归

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root.val < p.val && root.val < q.val){
            return lowestCommonAncestor(root.right,p,q);
        }
        if(root.val > p.val && root.val > q.val){
            return lowestCommonAncestor(root.left,p,q);
        }
        return root;
    }

}