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CyC2018
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@ -4260,16 +4260,16 @@ public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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}
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```
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**最大相同节点值的路径长度**
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**相同节点值的最大路径长度**
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[Leetcode : 687. Longest Univalue Path (Easy)](https://pomotodo.com/app/)
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[Leetcode : 687. Longest Univalue Path (Easy)](https://leetcode.com/problems/longest-univalue-path/)
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```html
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1
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1
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/ \
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4 5
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/ \ \
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4 4 5
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4 5
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/ \ \
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4 4 5
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Output : 2
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```
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@ -4296,6 +4296,15 @@ private int dfs(TreeNode root){
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[Leetcode : 337. House Robber III (Medium)](https://leetcode.com/problems/house-robber-iii/description/)
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```html
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3
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/ \
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2 3
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\ \
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3 1
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Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
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```
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```java
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public int rob(TreeNode root) {
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if (root == null) return 0;
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@ -4313,9 +4322,9 @@ public int rob(TreeNode root) {
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### 层次遍历
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使用 BFS，不需要使用两个队列来分别存储当前层的节点和下一层的节点， 因为在开始遍历一层的节点时，当前队列中的节点数就是当前层的节点数，只要控制遍历这么多节点数，就能保证这次遍历的都是当前层的节点。
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使用 BFS 进行层次遍历。不需要使用两个队列来分别存储当前层的节点和下一层的节点，因为在开始遍历一层的节点时，当前队列中的节点数就是当前层的节点数，只要控制遍历这么多节点数，就能保证这次遍历的都是当前层的节点。
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**计算一棵树每层节点的平均数**
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**一棵树每层节点的平均数**
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[637. Average of Levels in Binary Tree (Easy)](https://leetcode.com/problems/average-of-levels-in-binary-tree/description/)
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@ -4344,6 +4353,21 @@ public List<Double> averageOfLevels(TreeNode root) {
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[Leetcode : 513. Find Bottom Left Tree Value (Easy)](https://leetcode.com/problems/find-bottom-left-tree-value/description/)
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```html
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Input:
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1
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/ \
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2 3
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/ / \
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4 5 6
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/
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7
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Output:
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7
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```
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```java
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public int findBottomLeftValue(TreeNode root) {
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Queue<TreeNode> queue = new LinkedList<>();
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@ -4426,9 +4450,9 @@ public List<Integer> preorderTraversal(TreeNode root) {
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}
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```
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**非递归实现二叉树的后续遍历**
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**非递归实现二叉树的后序遍历**
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[Leetcode : ### 145. Binary Tree Postorder Traversal (Medium)](https://leetcode.com/problems/binary-tree-postorder-traversal/description/)
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[Leetcode : 145. Binary Tree Postorder Traversal (Medium)](https://leetcode.com/problems/binary-tree-postorder-traversal/description/)
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前序遍历为 root -> left -> right，后序遍历为 left -> right -> root，可以修改前序遍历成为 root -> right -> left，那么这个顺序就和后序遍历正好相反。
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@ -4471,15 +4495,26 @@ public List<Integer> inorderTraversal(TreeNode root) {
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}
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```
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**使用中序遍历和前序遍历序列重建二叉树** //TODO
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### BST
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主要利用 BST 中序遍历有序的特点。
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**在 BST 中寻找两个节点，使它们的和为一个给定值。**
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**在 BST 中寻找两个节点，使它们的和为一个给定值**
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[653. Two Sum IV - Input is a BST](https://leetcode.com/problems/two-sum-iv-input-is-a-bst/description/)
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[653. Two Sum IV - Input is a BST (Easy)](https://leetcode.com/problems/two-sum-iv-input-is-a-bst/description/)
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```html
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Input:
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5
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/ \
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3 6
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/ \ \
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2 4 7
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Target = 9
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Output: True
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```
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使用中序遍历得到有序数组之后，再利用双指针对数组进行查找。
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@ -4507,10 +4542,22 @@ private void inOrder(TreeNode root, List<Integer> nums){
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}
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```
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**在 BST 中查找最小的两个节点之差的绝对值**
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**在 BST 中查找两个节点之差的最小绝对值**
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[Leetcode : 530. Minimum Absolute Difference in BST (Easy)](https://leetcode.com/problems/minimum-absolute-difference-in-bst/description/)
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```html
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Input:
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1
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\
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3
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/
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2
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Output:
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1
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```
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利用 BST 的中序遍历为有序的性质，计算中序遍历中临近的两个节点之差的绝对值，取最小值。
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```java
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@ -4535,6 +4582,18 @@ private void inorder(TreeNode node){
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[Leetcode : Convert BST to Greater Tree (Easy)](https://leetcode.com/problems/convert-bst-to-greater-tree/description/)
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```html
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Input: The root of a Binary Search Tree like this:
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5
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/ \
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2 13
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Output: The root of a Greater Tree like this:
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18
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/ \
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20 13
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```
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先遍历右子树。
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```java
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@ -4562,6 +4621,17 @@ private void traver(TreeNode root) {
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**寻找 BST 中出现次数最多的节点**
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[Leetcode : 501. Find Mode in Binary Search Tree (Easy)](https://leetcode.com/problems/find-mode-in-binary-search-tree/description/)
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```html
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1
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\
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2
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/
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2
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return [2].
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```
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```java
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private int cnt = 1;
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private int maxCnt = 1;
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@ -4711,6 +4781,13 @@ class Trie {
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[Leetcode : 677. Map Sum Pairs (Medium)](https://leetcode.com/problems/map-sum-pairs/description/)
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```html
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Input: insert("apple", 3), Output: Null
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Input: sum("ap"), Output: 3
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Input: insert("app", 2), Output: Null
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Input: sum("ap"), Output: 5
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```
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```java
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class MapSum {
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private class Trie {
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