* };*/classSolution {public:intpreroot =0; TreeNode* bstToGst(TreeNode*root) {if(root->right) bstToGst(root->right); preroot= root->val = root->val +preroot;if(root->left) bstToGst(root->left);returnroot; } };
* } */classSolution{publicTreeNodebstToGst(TreeNode root){if(root ==null)returnnull;intsum=0; bstToGstInt(root,sum);returnroot; }publicintbstToGstInt(TreeNode node,intsum){if(node ==null)returnsum; sum = bstToGstInt(node.right,sum); sum += node.val; node.val = sum; sum = b...
var bstToGst = function(root) { let sum = 0 function traversal (root) { if (root !== null) { traversal(root.right) root.val += sum sum = root.val traversal(root.left) } } traversal(root) return root } 16、把二叉搜索树转换为累加树 LeetCode 第238题 var convertBST = function(...
# class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def bstToGst(self, root: TreeNode) -> TreeNode: self.val = 0 def helper(root): if not root: return helper(root.right) self.val += root.val root.val = ...
给定一个二叉搜索树root(BST),请将它的每个节点的值替换成树中大于或者等于该节点值的所有节点值之和。 提醒一下,二叉搜索树满足下列约束条件: 节点的左子树仅包含键小于节点键的节点。 节点的右子树仅包含键大于节点键的节点。 左右子树也必须是二叉搜索树。
class Solution { public final int maxn = 101; public int[] val_lst = new int[maxn]; public int cot = -1; public TreeNode bstToGst(TreeNode root) { look(root); sum(); cot = -1; return change(root); } public void look(TreeNode root){ if(root==null) return; look(root.lef...
() def top(self): """ Get the top element. :rtype: int """ return self.stack[-1] def empty(self): """ Returns whether the stack is empty. :rtype: bool """ return True if not self.stack else False # %% # * Two Sum IV - Input is a BST def findTarget(self, root, k...
No_0538_Convert BST to Greater Tree No_0540_Single Element in a Sorted Array No_0541_Reverse String II No_0543_Diameter of Binary Tree No_0554_Brick Wall No_0557_Reverse Words in a String III No_0559_Maximum Depth of N-ary Tree No_0561_Array Partition I No_0563_Binary Tree...
func bstToGst(root *TreeNode) *TreeNode { nums:=helper(root) mnums:=make(map[int]int) length:=len(nums) for i:=0;i<length;i++{ mnums[nums[i]]=i } return solve(root,nums,mnums) } func helper(root *TreeNode)[]int{//存放节点 ...
classSolution{public:intans =0;TreeNode*bstToGst(TreeNode* root){if(root==nullptr)returnroot;bstToGst(root->right); ans+=root->val; root->val=ans;bstToGst(root->left);returnroot; } }; 多边形三角剖分的最低得分 题目:多边形三角剖分的最低得分(Minimum Score Triangulation of Polygon) ...