剑指 Offer 68 - II. 二叉树的最近公共祖先

题目链接：
剑指 Offer 68 - II. 二叉树的最近公共祖先 - 力扣（LeetCode） (leetcode-cn.com)

方法1(易理解，代码量大)：

//方法1：用一个HashMap存入所有节点对应的父节点(建立映射关系)，然后去遍历其中一个节点的所有父节点即
    //(包括父节点的父节点等)
    
    public static TreeNode lowestCommonAncestor1(TreeNode root,TreeNode p,TreeNode q) {
        HashMap<TreeNode,TreeNode> fatherMap = new HashMap<>();
        fatherMap.put(root, root);
        process1(root,fatherMap);
        HashSet<TreeNode> set = new HashSet<>();
        set.add(root);
        TreeNode cur = p;
        while (cur != fatherMap.get(cur)) {
            set.add(cur);
            cur = fatherMap.get(cur);
        }
        cur = q;
        while (cur != fatherMap.get(cur)) {
            if (set.contains(cur) == true) {
                return cur;
            }
            cur = fatherMap.get(cur);
        }
        return cur;
    }
    
    public static void process1(TreeNode root,HashMap<TreeNode,TreeNode> fatherMap) {
        if (root == null) {
            return;
        }
        fatherMap.put(root.left, root);
        fatherMap.put(root.right, root);
        process1(root.left,fatherMap);
        process1(root.right,fatherMap);
    }

 

方法2(难理解，代码简洁)：

//方法2：
    public static TreeNode lowestCommonAncestor(TreeNode root,TreeNode p,TreeNode q) {
        if (root == null || root == p || root == q) {//保证返回值只要null,p,q三者
            return root;
        }
        TreeNode left = lowestCommonAncestor(root.left,p,q);//找其左子树是否含有p或q
        TreeNode right = lowestCommonAncestor(root.right,p,q);//找其右子树是否含有p或q
        if (left != null && right != null) {//如果从root节点的左右子树都含有p,q,则第一次出现这种情况的节点
                                            //就是所要的节点,直接return root;得到结果
            return root;
        }
        return left != null ? left : right;//有p,q优先return p,q;如果都没有就return null
    }