2096. Step-By-Step Directions From a Binary Tree Node to Another

方法

先找到最近公共祖先，然后分别向下dfs，直到遇到dest或者source，同时记录路径。再把路径反过来，拼接。

class Solution {
    StringBuffer ps = new StringBuffer(); // 到source的路径
    StringBuffer pd = new StringBuffer(); // 到dest的路径
    StringBuffer res = new StringBuffer();

    public String getDirections(TreeNode root, int startValue, int destValue) {
        TreeNode lca = getLCA(root, startValue, destValue);
        pathToSrc(lca, startValue);
        pathToDest(lca, destValue);
        return res.toString();
    }
    
    // 找从lca到source的路径
    public void pathToSrc(TreeNode node, int val) {
        if (node == null) {
            return;
        }
        if (node.val == val) {
            res.append(ps); // no need to reverse, records "U"s
            return;
        }
        ps.append("U");
        pathToSrc(node.left, val);
        ps.deleteCharAt(ps.length() - 1);
        ps.append("U");
        pathToSrc(node.right, val);
        ps.deleteCharAt(ps.length() - 1);
    }

    // 找从lca到dest的路径
    public void pathToDest(TreeNode node, int val) {
        if (node == null) {
            return;
        }
        if (node.val == val) {
            res.append(pd);
            return;
        }
        pd.append("L");
        pathToDest(node.left, val);
        pd.deleteCharAt(pd.length() - 1);
        pd.append("R");
        pathToDest(node.right, val);
        pd.deleteCharAt(pd.length() - 1);
    }

    // 找出两个数的最近公共祖先
    public TreeNode getLCA(TreeNode root, int val1, int val2) {
        if (root == null || root.val == val1 || root.val == val2) {
            return root;
        }
        TreeNode left = getLCA(root.left, val1, val2);
        TreeNode right = getLCA(root.right, val1, val2);
        if (left != null && right != null) {
            return root;
        } else if (left != null & right == null) {
            return left;
        } else if (left == null & right != null) {
            return right;
        } else {
            return null;
        }

    }

}