因为,没有重复值,所以只需要做一个标记就OK了. public class Successor { static boolean flag = false; static int result = 0; public int findSucc(TreeNode root, int p) { // write code here inOrder(root,p); return result; } public static void inOrder(TreeNode root,int p){ if…
Given a binary search tree and a node in it, find the in-order successor of that node in the BST. The successor of a node p is the node with the smallest key greater than p.val. You will have direct access to the node but not to the root of the tree.…
Given a binary search tree and a node in it, find the in-order successor of that node in the BST. Note: If the given node has no in-order successor in the tree, return null. 这道题让我们求二叉搜索树的某个节点的中序后继节点,那么我们根据BST的性质知道其中序遍历的结果是有序的, 是我最先用的方法是用迭代的中序遍历方法,然后用…
Given a binary tree, return the inorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,3,2] Follow up: Recursive solution is trivial, could you do it iteratively? 题意: 二叉树中序遍历 Solution1: Recursion code class Soluti…
Given a binary search tree and a node in it, find the in-order successor of that node in the BST. The successor of a node p is the node with the smallest key greater than p.val. Example 1: Input: root = [2,1,3], p = 1 Output: 2 Explanation: 1's in-or…
