LeetCode 235 Lowest Common Ancestor of a Binary Search Tree

方法一： 坐等左边右边来送结果～～～，适合所有二叉树

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) {
            return null;
        }
        if (root == p || root == q) {
            return root;
        }

        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        if (left != null && right != null) {
            return root;
        }
        if (left != null) {
            return left;
        }
        if (right != null) {
            return right;
        }
        return null;
    }
}

方法二 Branch pruning : 

转来自老铁的博客https://home.cnblogs.com/u/davidnyc

 // the goal is to find the root that would sit in the middle
     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
         //base case
         if (root == null) return null ;
         int min = Math.min(p.val, q.val) ;
         int max = Math.max(p.val, q.val) ;
         //pre order：看看当前点是不是满足的值
         if (root.val>=min && root.val <= max) return root ;
         //the left: branch pruning > max, then go left： 左边返回就一层层往上返回
         if (root.val > max){
            return lowestCommonAncestor(root.left, p, q) ;
         }
         //the right: branch pruning: <min: then go right： 右边返回就一层层往上返回
         if (root.val < min){
           return lowestCommonAncestor(root.right, p, q) ;
         }
         //不从左边，也不从右边的话，就是NULL
         return null;
     }