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include/s0538_convert_bst_to_greater_tree.hpp
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include/s0538_convert_bst_to_greater_tree.hpp
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#ifndef S0538_CONVERT_BST_TO_GREATER_TREE_HPP
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#define S0538_CONVERT_BST_TO_GREATER_TREE_HPP
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#include "structures.hpp"
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class S0538 {
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public:
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int sum{0};
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TreeNode* convertBST1(TreeNode* root);
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TreeNode* convertBST2(TreeNode* root);
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};
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#endif
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- [s0450](https://leetcode.cn/problems/delete-node-in-a-bst/description/): 删除节点。递归删除。
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- [s0669](https://leetcode.cn/problems/trim-a-binary-search-tree/description/): 修剪 BST 。递归修剪。
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- [s0108](https://leetcode.cn/problems/convert-sorted-array-to-binary-search-tree/): 有序数组转 BST 。数组中点为根节点,中点左侧部分生成左子树,右侧部分生成右子树,递归。
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- [s0538](https://leetcode.cn/problems/convert-bst-to-greater-tree/description/): BST 转累加树。可以找到每个节点的构建方法然后用直观一点的递归方式来写,不过本题有个特殊之处在于累加树的生成方式正好和反序中序遍历的遍历路径相同,因此可以用反序中序遍历来遍历生成。
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src/s0538_convert_bst_to_greater_tree.cpp
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src/s0538_convert_bst_to_greater_tree.cpp
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#include "s0538_convert_bst_to_greater_tree.hpp"
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// 思路一:直观递归
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TreeNode* S0538::convertBST1(TreeNode* root) {
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// 为空则直接返回
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if (root == nullptr) return nullptr;
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// 更新右子树
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root->right = convertBST1(root->right);
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// 更新根节点的值
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if (root->right != nullptr) {
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// 找到右子树的最左侧节点
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TreeNode* node = root->right;
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while (node->left != nullptr) node = node->left;
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// 根节点 = 根节点 + 右子树的最左侧节点
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root->val += node->val;
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}
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// 接下来处理左子树
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if (root->left != nullptr) {
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// 先把根节点的值加到左子树的最右侧点
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TreeNode* node = root->left;
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// 找到左子树的最右侧点
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while (node->right != nullptr) node = node->right;
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// 加上去
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node->val += root->val;
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// 更新左子树
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root->left = convertBST1(root->left);
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}
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// 返回
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return root;
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}
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// 思路二:反序中序遍历
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// 反序中序遍历的遍历路径正好和累加树的路径相同,因此可以直接用它来遍历
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// 用一个全局变量 sum 记录遍历过程中每个节点的值
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TreeNode* S0538::convertBST2(TreeNode* root) {
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if (root != nullptr) {
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convertBST2(root->right);
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sum += root->val;
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root->val = sum;
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convertBST2(root->left);
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}
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return root;
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}
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