You are given therootof a binary search tree (BST), where the values ofexactlytwo nodes of the tree were swapped by mistake.Recover the tree without changing its structure. Example 1: Input:root = [1,3,null,null,2]Output:[3,1,null,null,2]Explanation:3 cannot be a left child of 1...
Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/recover-binary-search-tree 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 【题意分析】 对于一个B...
1/**2* Definition for binary tree3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode *right;7* TreeNode(int x) : val(x), left(NULL), right(NULL) {}8* };9*/10classSolution {11public:12voidrecoverTree(TreeNode *root) {13n1 = n2 = pre =NULL;14inorder(root);15...
* type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */ func recoverTree(root *TreeNode) { // 定义 first 和 second ,用于维护交换的两个结点 // previous: 表示中序遍历的前一个结点 // first: 表示互换结点的前者,该结点必定比后一个结点大 var first, second *Tree...
Recover Binary Search Tree -- LeetCode http://oj.leetcode.com/problems/recover-binary-search-tree/ 这道题是要求恢复一颗有两个元素调换错了的二叉查找树。一開始拿到可能会认为比較复杂,事实上观察出规律了就比較简单。主要还是利用二叉查找树的主要性质,就是中序遍历是有序的性质。那么假设当中有元素被调换...
packageleetcodeimport("/halfrost/LeetCode-Go/structures")// TreeNode definetypeTreeNode=structures.TreeNode/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */funcrecoverTree(root*TreeNode){varprev,target1,target2*TreeNo...
publicvoidrecoverTree2(TreeNoderoot){if(root==null){return;}//寻找左子树中最大的节点TreeNodemaxLeft=getMaxOfBST(root.left);//寻找右子树中最小的节点TreeNodeminRight=getMinOfBST(root.right);if(minRight!=null&&maxLeft!=null){//左边的大于根节点,右边的小于根节点,对应情况 3,左右子树中的两...
* Definition for a binary tree node.* struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void recoverTree(TreeNode* root) { ...
Can you solve this real interview question? Recover a Tree From Preorder Traversal - We run a preorder depth-first search (DFS) on the root of a binary tree. At each node in this traversal, we output D dashes (where D is the depth of this node), then we
TBC Time Complexity: O(N) Space Complexity: O(1) Solution1 Code: classSolution{privateTreeNode node1,node2;privateTreeNode prev=newTreeNode(Integer.MIN_VALUE);publicvoidrecoverTree(TreeNode root){if(root==null)return;inorderCheck(root);int tmp=node1.val;node1.val=node2.val;node2.val=tm...