[LC] 426. Convert Binary Search Tree to Sorted Doubly Linked List

原创于 2019-12-15 12:22:09 发布 · 161 阅读

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本文介绍了一种将二叉搜索树(BST)转换为双向链表的方法，提供了递归和使用Stack的两种解决方案。通过Inorder DFS遍历，确保了节点按排序顺序连接。

BST换排好序的任何数据结构用脚趾头想都应该是inorder的dfs。这里也就是dfs中间穿插一些前后连接的操作就行了。
给出两种解法，一个是递归做的，一个是用Stack做的。（意想不到的是递归的比Stack的要快）

    public Node treeToDoublyList(Node root) {
        if (root == null) return null;

        Node[] prev = new Node[1];
        Node[] newRoot = new Node[1];
        dfs(root, prev, newRoot);
        newRoot[0].left = prev[0];
        prev[0].right = newRoot[0];
        return newRoot[0];
    }
    
    public void dfs(Node current, Node[] prev, Node[] newRoot) {
        if (current != null) {
            dfs(current.left, prev, newRoot);
            if (newRoot[0] == null) {
                newRoot[0] = current;
            }
            
            if (prev[0] != null) {
                prev[0].right = current;
                current.left = prev[0];
            }
            
            prev[0] = current;
            dfs(current.right, prev, newRoot);
        }
    }

    public Node treeToDoublyList(Node root) {
        if (root == null) return null;

        Node current = root;
        Node prev = null, firstNode = null;
        Stack<Node> nodeStk = new Stack<>();
        while(current != null || !nodeStk.isEmpty()) {
            if (current != null) {
                nodeStk.push(current);
                current = current.left;
            } else {
                current = nodeStk.pop();
                if (firstNode == null) {
                    firstNode = current;
                }

                if (prev != null) {
                    prev.right = current;
                    current.left = prev;
                }
                prev = current;
                current = current.right;
            }
        }
        prev.right = firstNode;
        firstNode.left = prev;
        return firstNode;
    }