leetcode236 - Lowest Common Ancestor of a Binary Tree - medium

class Solution {
    public static TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) {
            return null;
        }
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);

        if (root == p || root == q) {
            boolean rootright = true;
            return root;
        }
        if (left != null && right != null) {
            boolean notempty = true;
            return root;
        }

        return right == null ? left : right;
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private class ResultType{
        boolean hasP;
        boolean hasQ;
        TreeNode lca;
        public ResultType(boolean hasP, boolean hasQ, TreeNode lca) {
            this.hasP = hasP;
            this.hasQ = hasQ;
            this.lca = lca;
        }
    }
    
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        ResultType rt = helper(root, p, q);
        return rt.lca;
    }
    
    private ResultType helper(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) {
            return new ResultType(false, false, null);
        }
        // if (root == p) {
        //     return new ResultType(true, false, null);
        // }
        // if (root == q) {
        //     return new ResultType(false, true, null);
        // }
        // P1: 一个是不要处理的这么细，拿root和pq是否相等也这样处理，尽可能包含在下面的递归里。另一个是hasP本来就要考虑当前root是不是的情况，如果左右都没有p但本root就是p那也得是hasP啊。这样才能正确处理到p是q的祖先的情况。
        ResultType left = helper(root.left, p, q);
        ResultType right = helper(root.right, p, q);
        boolean hasP = left.hasP || right.hasP || root == p;
        boolean hasQ = left.hasQ || right.hasQ || root == q;
        TreeNode lca = null;
        if (hasP && hasQ) {
            if (left.lca != null) {
                lca = left.lca;
            } else if (right.lca != null) {
                lca = right.lca;
            } else {
                lca = root;
            }
        }
        return new ResultType(hasP, hasQ, lca);
    }
}