输入一棵二叉搜索树,将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的结点,只能调整树中结点指针的指向。
代码
static class TreeNode { int val; TreeNode left; TreeNode right; public TreeNode(int val) { this.val = val; } } public static TreeNode convertTree2LinkedList(TreeNode root) { if (root == null) { return null; } TreeNode head = convert(root); // 找到链表的头节点,即指向前一节点的left为null while (head.left != null) { head = head.left; } return head; } /** * 采用递归的方式 * 在双向链表中,left节点指向前一节点,right指向下一节点 * @param root * @return */ private static TreeNode convert(TreeNode root) { if (root == null) { return null; } // 如果存在左子树,则将左子树转换为双向链表 if (root.left != null) { // 左子树双向链表 TreeNode left = convertTree2LinkedList(root.left); // 因为在双向链表中right指向后一节点,找到右子节点为null的节点,作为左子树双向链表的尾节点 while (left.right != null) { left = left.right; } // 将树的根节点放到左子树双向链表的尾节点的后面 root.left = left; left.right = root; } // 如果存在右子树,则将右子树转换为双向链表 if (root.right != null) { // 右子树双向链表 TreeNode right = convertTree2LinkedList(root.right); // 因为在双向链表中left指向前一节点,找到左子节点为null的节点,作为右子树双向链表的头节点 while (right.left != null) { right = right.left; } // 将右子树双向链表的头节点放到树的根节点的后面 root.right = right; right.left = root; } // 递归终止条件,没有左右子节点的节点时返回 return root; } public static void main(String[] args) { TreeNode root = buildTree(); root = convertTree2LinkedList(root); // 输出 2 3 4 5 7 while (root != null) { System.out.print(root.val + " "); root = root.right; } } /** * 创建tree: * 5 * 3 7 * 2 4 * @return */ private static TreeNode buildTree(){ TreeNode root = new TreeNode(5); TreeNode left = new TreeNode(3); TreeNode left1 = new TreeNode(2); TreeNode right1 = new TreeNode(4); left.left = left1; left.right = right1; TreeNode right = new TreeNode(7); root.left = left; root.right = right; return root; }