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CyC2018
@ -477,7 +477,7 @@ System.out.println(InterfaceExample.x);
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- 从设计层面上看，抽象类提供了一种 IS-A 关系，那么就必须满足里式替换原则，即子类对象必须能够替换掉所有父类对象。而接口更像是一种 LIKE-A 关系，它只是提供一种方法实现契约，并不要求接口和实现接口的类具有 IS-A 关系。
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- 从使用上来看，一个类可以实现多个接口，但是不能继承多个抽象类。
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- 接口的字段只能是 static 和 final 类型的，而抽象类的字段没有这种限制。
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- 接口的方法只能是 public 的，而抽象类的方法可以由多种访问权限。
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- 接口的方法只能是 public 的，而抽象类的方法可以有多种访问权限。
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**4. 使用选择**
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@ -5728,10 +5728,10 @@ public List<Double> averageOfLevels(TreeNode root) {
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if (root == null) return ret;
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Queue<TreeNode> queue = new LinkedList<>();
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queue.add(root);
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while (!queue.isEmpty()){
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while (!queue.isEmpty()) {
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int cnt = queue.size();
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double sum = 0;
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for (int i = 0; i < cnt; i++){
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for (int i = 0; i < cnt; i++) {
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TreeNode node = queue.poll();
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sum += node.val;
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if (node.left != null) queue.add(node.left);
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@ -5766,7 +5766,7 @@ Output:
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public int findBottomLeftValue(TreeNode root) {
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Queue<TreeNode> queue = new LinkedList<>();
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queue.add(root);
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while (!queue.isEmpty()){
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while (!queue.isEmpty()) {
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root = queue.poll();
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if (root.right != null) queue.add(root.right);
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if (root.left != null) queue.add(root.left);
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@ -5785,10 +5785,10 @@ public int findBottomLeftValue(TreeNode root) {
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4 5 6
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```
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层次遍历顺序：[1 2 3 4 5 6]
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前序遍历顺序：[1 2 4 5 3 6]
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中序遍历顺序：[4 2 5 1 3 6]
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后序遍历顺序：[4 5 2 6 3 1]
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- 层次遍历顺序：[1 2 3 4 5 6]
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- 前序遍历顺序：[1 2 4 5 3 6]
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- 中序遍历顺序：[4 2 5 1 3 6]
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- 后序遍历顺序：[4 5 2 6 3 1]
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层次遍历使用 BFS 实现，利用的就是 BFS 一层一层遍历的特性；而前序、中序、后序遍历利用了 DFS 实现。
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@ -5797,7 +5797,7 @@ public int findBottomLeftValue(TreeNode root) {
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① 前序
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```java
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void dfs(TreeNode root){
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void dfs(TreeNode root) {
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visit(root);
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dfs(root.left);
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dfs(root.right);
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@ -5807,7 +5807,7 @@ void dfs(TreeNode root){
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② 中序
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```java
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void dfs(TreeNode root){
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void dfs(TreeNode root) {
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dfs(root.left);
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visit(root);
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dfs(root.right);
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@ -5817,7 +5817,7 @@ void dfs(TreeNode root){
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③ 后序
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```java
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void dfs(TreeNode root){
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void dfs(TreeNode root) {
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dfs(root.left);
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dfs(root.right);
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visit(root);
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@ -5837,7 +5837,7 @@ public List<Integer> preorderTraversal(TreeNode root) {
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TreeNode node = stack.pop();
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if (node == null) continue;
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ret.add(node.val);
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stack.push(node.right); // 先右后左，保证左子树先遍历
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stack.push(node.right); // 先右后左，保证左子树先遍历
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stack.push(node.left);
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}
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return ret;
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@ -5923,7 +5923,7 @@ Output:
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1
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```
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只保留值在 L \~ R 之间的节点
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题目描述：只保留值在 L \~ R 之间的节点
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```java
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public TreeNode trimBST(TreeNode root, int L, int R) {
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@ -6008,12 +6008,12 @@ public TreeNode convertBST(TreeNode root) {
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return root;
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}
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private void traver(TreeNode root) {
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if (root == null) return;
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if (root.right != null) traver(root.right);
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sum += root.val;
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root.val = sum;
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if (root.left != null) traver(root.left);
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private void traver(TreeNode node) {
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if (node == null) return;
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traver(node.right);
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sum += node.val;
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node.val = sum;
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traver(node.left);
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}
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```
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@ -6085,7 +6085,6 @@ private TreeNode toBST(int[] nums, int sIdx, int eIdx){
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}
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```
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**根据有序链表构造平衡的二叉查找树**
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[109. Convert Sorted List to Binary Search Tree (Medium)](https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/description/)
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@ -6105,29 +6104,25 @@ One possible answer is: [0,-3,9,-10,null,5], which represents the following heig
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```java
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public TreeNode sortedListToBST(ListNode head) {
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if (head == null) return null;
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int size = size(head);
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if (size == 1) return new TreeNode(head.val);
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ListNode pre = head, mid = pre.next;
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int step = 2;
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while (step <= size / 2) {
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pre = mid;
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mid = mid.next;
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step++;
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}
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pre.next = null;
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if (head.next == null) return new TreeNode(head.val);
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ListNode preMid = preMid(head);
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ListNode mid = preMid.next;
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preMid.next = null; // 断开链表
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TreeNode t = new TreeNode(mid.val);
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t.left = sortedListToBST(head);
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t.right = sortedListToBST(mid.next);
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return t;
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}
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private int size(ListNode node) {
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int size = 0;
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while (node != null) {
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size++;
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node = node.next;
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private ListNode preMid(ListNode head) {
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ListNode slow = head, fast = head.next;
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ListNode pre = head;
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while (fast != null && fast.next != null) {
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pre = slow;
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slow = slow.next;
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fast = fast.next.next;
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}
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return size;
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return pre;
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}
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```
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@ -6207,7 +6202,7 @@ public int getMinimumDifference(TreeNode root) {
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private void inOrder(TreeNode node) {
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if (node == null) return;
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inOrder(node.left);
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if (preNode != null) minDiff = Math.min(minDiff, Math.abs(node.val - preNode.val));
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if (preNode != null) minDiff = Math.min(minDiff, node.val - preNode.val);
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preNode = node;
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inOrder(node.right);
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}
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