669. 修剪二叉搜索树

由于递归函数有返回值，对于不在范围内的节点，可以通过左右孩子递归来实现修剪。

接下来要将下一层处理完左子树的结果赋给root->left，处理完右子树的结果赋给root->right。

最后返回root节点。

class Solution {public TreeNode trimBST(TreeNode root, int low, int high) {if (root == null) return null;if (root.val > high) return trimBST(root.left, low,high);if (root.val <  low) return trimBST(root.right,low,high);root.left = trimBST(root.left, low,high);root.right= trimBST(root.right,low,high);return root;}
}

108.将有序数组转换为二叉搜索树

吸取了之前构造二叉树的经验，递归函数需要利用开始索引和结束索引

注意这里二分采用左闭右开的写法，因此调用递归函数右边是nums.length

class Solution {public TreeNode sortedArrayToBST(int[] nums) {return BST(nums, 0, nums.length);}TreeNode BST(int[] nums, int left, int right) {if (left >= right) return null;if (right-left == 1) return new TreeNode(nums[left]);TreeNode root = new TreeNode(nums[(right+left)/2]);root.left = BST(nums,left, (right+left)/2);root.right= BST(nums,(right+left)/2+1,right);return root;}
}

538.把二叉搜索树转换为累加树

直接暴力做了，先求到总和，再按照中序遍历构建

class Solution {int sum = 0;public TreeNode convertBST(TreeNode root) {getSum(root);addTree(root);return root;}void getSum(TreeNode root) {if (root == null) return;sum += root.val;getSum(root.left);getSum(root.right);}void addTree(TreeNode root) {if (root == null) return ;addTree(root.left);int temp = root.val;root.val = sum;sum -= temp;addTree(root.right);}
}

看题解发现按照右中左顺序来遍历就可以了

class Solution {int sum;public TreeNode convertBST(TreeNode root) {sum = 0;convertBST1(root);return root;}public void convertBST1(TreeNode root) {if (root == null) return;convertBST1(root.right);sum += root.val;root.val = sum;convertBST1(root.left);}
}