代码随想录第二十三天-拓冰建站

第六章二叉树part09

今日内容：

● 669. 修剪二叉搜索树

● 108.将有序数组转换为二叉搜索树

● 538.把二叉搜索树转换为累加树

● 总结篇

详细布置

669. 修剪二叉搜索树

这道题目比较难，比添加增加和删除节点难的多，建议先看视频理解。

题目链接/文章讲解： https://programmercarl.com/0669.%E4%BF%AE%E5%89%AA%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91.html

视频讲解： https://www.bilibili.com/video/BV17P41177ud

思路：

步骤：

完整代码

递归

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

迭代

class Solution {//iterationpublic TreeNode trimBST(TreeNode root, int low, int high) {if(root == null)return null;while(root != null && (root.val < low || root.val > high)){if(root.val < low)root = root.right;elseroot = root.left;}TreeNode curr = root;//deal with root's left sub-tree, and deal with the value smaller than low.while(curr != null){while(curr.left != null && curr.left.val < low){curr.left = curr.left.right;}curr = curr.left;}//go back to root;curr = root;//deal with root's righg sub-tree, and deal with the value bigger than high.while(curr != null){while(curr.right != null && curr.right.val > high){curr.right = curr.right.left;}curr = curr.right;}return root;}
}

108.将有序数组转换为二叉搜索树

本题就简单一些，可以尝试先自己做做。

https://programmercarl.com/0108.%E5%B0%86%E6%9C%89%E5%BA%8F%E6%95%B0%E7%BB%84%E8%BD%AC%E6%8D%A2%E4%B8%BA%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91.html

视频讲解：https://www.bilibili.com/video/BV1uR4y1X7qL

思路：

步骤：

完整代码

递归: 左闭右开 [left,right)

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

递归: 左闭右闭 [left,right]

class Solution {public TreeNode sortedArrayToBST(int[] nums) {TreeNode root = traversal(nums, 0, nums.length - 1);return root;}// 左闭右闭区间[left, right]private TreeNode traversal(int[] nums, int left, int right) {if (left > right) return null;int mid = left + ((right - left) >> 1);TreeNode root = new TreeNode(nums[mid]);root.left = traversal(nums, left, mid - 1);root.right = traversal(nums, mid + 1, right);return root;}
}

迭代: 左闭右闭 [left,right]

class Solution {public TreeNode sortedArrayToBST(int[] nums) {if (nums.length == 0) return null;//根节点初始化TreeNode root = new TreeNode(-1);Queue<TreeNode> nodeQueue = new LinkedList<>();Queue<Integer> leftQueue = new LinkedList<>();Queue<Integer> rightQueue = new LinkedList<>();// 根节点入队列nodeQueue.offer(root);// 0为左区间下标初始位置leftQueue.offer(0);// nums.size() - 1为右区间下标初始位置rightQueue.offer(nums.length - 1);while (!nodeQueue.isEmpty()) {TreeNode currNode = nodeQueue.poll();int left = leftQueue.poll();int right = rightQueue.poll();int mid = left + ((right - left) >> 1);// 将mid对应的元素给中间节点currNode.val = nums[mid];// 处理左区间if (left <= mid - 1) {currNode.left = new TreeNode(-1);nodeQueue.offer(currNode.left);leftQueue.offer(left);rightQueue.offer(mid - 1);}// 处理右区间if (right >= mid + 1) {currNode.right = new TreeNode(-1);nodeQueue.offer(currNode.right);leftQueue.offer(mid + 1);rightQueue.offer(right);}}return root;}
}

538.把二叉搜索树转换为累加树

本题也不难，在求二叉搜索树的最小绝对差和众数那两道题目都讲过了双指针法，思路是一样的。

https://programmercarl.com/0538.%E6%8A%8A%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E8%BD%AC%E6%8D%A2%E4%B8%BA%E7%B4%AF%E5%8A%A0%E6%A0%91.html

视频讲解：https://www.bilibili.com/video/BV1d44y1f7wP

思路：

步骤：

完整代码

递归

class Solution {int sum;public TreeNode convertBST(TreeNode root) {sum = 0;convertBST1(root);return root;}// 按右中左顺序遍历，累加即可public void convertBST1(TreeNode root) {if (root == null) {return;}convertBST1(root.right);sum += root.val;root.val = sum;convertBST1(root.left);}
}

迭代

class Solution {//DFS iteraion統一迭代法public TreeNode convertBST(TreeNode root) {int pre = 0;Stack<TreeNode> stack = new Stack<>();if(root == null) //edge case checkreturn null;stack.add(root);while(!stack.isEmpty()){TreeNode curr = stack.peek();//curr != null的狀況，只負責存node到stack中if(curr != null){ stack.pop();if(curr.left != null)       //左stack.add(curr.left);stack.add(curr);            //中stack.add(null);if(curr.right != null)      //右stack.add(curr.right);}else{//curr == null的狀況，只負責做單層邏輯stack.pop();TreeNode temp = stack.pop();temp.val += pre;pre = temp.val;}}return root;}
}