Tree
Key points
Traversal:
pre-order, in-order, post-order
Two approaches: 1) DFS, 2) Iteration
Divide & Conquer VS Traversal:
DC: one send left & right two people to get the result, and then do sth.
T: one with a notebook walk through the tree
BST, Binary Search Tree:
in-order is a sorted list, important !
left < root < right, not duplicate
use
TreeSetorTreeMap
Tricks:
isLeaf(): root.left == null && root.right == null
every node: 1)root == null 2) root.left == null 3)root.right == null
Binary Indexed Tree:
the sum of (0, i), in O(logn) time.
Segment Tree:
sum/minimum in a range
Traverse
Pre-order
https://leetcode.com/problems/binary-tree-preorder-traversal/
Iterative
public class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
Stack<TreeNode> stack = new Stack<TreeNode>();
List<Integer> res = new LinkedList<Integer>();
TreeNode curr = root;
while (curr != null || !stack.isEmpty()) {
while (curr != null) {
res.add(curr.val);
stack.push(curr);
curr = curr.left;
}
TreeNode head = stack.pop();
curr = head.right;
}
return res;
}
}Recursive
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new LinkedList<Integer>();
traverse(root, res);
return res;
}
private void traverse(TreeNode root, List<Integer> res) {
if (root == null) return;
res.add(root.val);
traverse(root.left, res);
traverse(root.right, res);
}
}In-order
https://leetcode.com/problems/binary-tree-inorder-traversal/
Iterative
public class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
Stack<TreeNode> stack = new Stack<TreeNode>();
List<Integer> res = new LinkedList<Integer>();
TreeNode curr = root;
while (curr != null || !stack.isEmpty()) {
while (curr != null) {
stack.push(curr);
curr = curr.left;
}
TreeNode head = stack.pop();
res.add(head.val);
curr = head.right;
}
return res;
}
}Recursive
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new LinkedList<Integer>();
traverse(root, res);
return res;
}
private void traverse(TreeNode root, List<Integer> res) {
if (root == null) return;
traverse(root.left, res);
res.add(root.val);
traverse(root.right, res);
}
}Post-order
https://leetcode.com/problems/binary-tree-postorder-traversal/
Iterative
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
Stack<TreeNode> stack = new Stack<TreeNode>();
LinkedList<Integer> res = new LinkedList<Integer>();
TreeNode curr = root;
while (curr != null || !stack.isEmpty()) {
while (curr != null) {
res.addFirst(curr.val); // Revered of pre-order
stack.push(curr);
curr = curr.right; // Revered of pre-order
}
TreeNode head = stack.pop();
curr = head.left; // Revered of pre-order
}
return res;
}
}Recursive
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> res = new LinkedList<Integer>();
traverse(root, res);
return res;
}
private void traverse(TreeNode root, List<Integer> res) {
if (root == null) return;
traverse(root.left, res);
traverse(root.right, res);
res.add(root.val);
}
}Reference: https://discuss.leetcode.com/topic/30632/preorder-inorder-and-postorder-iteratively-summarization
Serialization and deserialization
https://leetcode.com/problems/find-duplicate-subtrees https://leetcode.com/problems/serialize-and-deserialize-binary-tree/description/ https://leetcode.com/problems/serialize-and-deserialize-bst https://leetcode.com/problems/construct-string-from-binary-tree https://leetcode.com/problems/construct-binary-tree-from-string
DFS or BFS?
BFS: level
DFS: go through a path
Common law
性质1：二叉树第i层上的结点数目最多为 2{i-1} (i≥1)。
性质2：深度为k的二叉树至多有2{k}-1个结点(k≥1)。
性质3：包含n个结点的二叉树的高度至少为log2 (n+1)。
性质4：在任意一棵二叉树中，若终端结点的个数为n0，度为2的结点数为n2，则n0=n2+1。Array -> Complete Binary Tree: parent i, left 2i+1, right 2i+2
Types of Binary Tree
Full Binary Tree
Complete Binary Tree
Binary Search Tree:
left < root < rightuse
TreeSetto implement BST: https://leetcode.com/problems/contains-duplicate-iii.TreeSetis implemented by Black-Red Tree, self balanced.insert,delete,searchinO(logN)complexity.
AVL Tree:
|left height - right height| <= 1
Heap: Max Heap, Min Heap. Heap is complete binary tree.
Priority Queue is a Heap, implemented using an Array object[]. More details about heapify: https://jaywinhuang.gitbooks.io/leetcode/content/mySolutions/023-merge-k-sorted-lists.html 
Segment Tree
the leaves are values of nums
https://leetcode.com/problems/range-sum-query-mutable/description/
ac2: Segment Tree
more intuitive, yet more code. Notice: use divide and conquer, no need to care about the mid point unless building the tree.
class NumArray {
class SegTreeNode{
int start, end, sum;
SegTreeNode left, right;
public SegTreeNode(int start, int end){
this.start = start;
this.end = end;
sum = 0;
}
}
SegTreeNode root;
public NumArray(int[] nums) {
root = build(nums, 0, nums.length-1);
}
private SegTreeNode build(int[] nums, int start, int end) {
if (start < 0 || end >= nums.length || start > end) return null;
int mid = start + (end - start) / 2;
SegTreeNode node = new SegTreeNode(start, end);
if (start == end) {
node.sum = nums[start];
} else {
node.left = build(nums, start, mid);
node.right = build(nums, mid+1, end);
node.sum = node.left.sum + node.right.sum;
}
return node;
}
public void update(int i, int val) {
update(root, i, val);
}
private void update(SegTreeNode root, int i, int val) {
if (i < root.start || i > root.end) return; // out of range
if (root.start == root.end) {
root.sum = val; // leaf, sum = nums[i]
return;
}
// divide and conquer
update(root.left, i, val);
update(root.right, i, val);
root.sum = root.left.sum + root.right.sum;
}
public int sumRange(int i, int j) {
return sumRange(root, i, j);
}
private int sumRange(SegTreeNode root, int i, int j) {
if (i > root.end || j < root.start) return 0; // out of range
if (i <= root.start && root.end <= j) return root.sum; // current node within range, return value
// divide and conquer
int res = 0;
res += sumRange(root.left, i, j);
res += sumRange(root.right, i, j);
return res;
}
}
/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* obj.update(i,val);
* int param_2 = obj.sumRange(i,j);
*/Binary Indexed Tree
https://leetcode.com/problems/range-sum-query-mutable/description/ https://leetcode.com/problems/range-sum-query-2d-mutable/description/
ac1: Binary Indexed Tree
careful: bit.length = nums.length + 1; so need to do i++ in methods
class NumArray {
int[] bit;
int[] nums;
public NumArray(int[] nums) {
this.nums = nums;
bit = new int[nums.length + 1];
for (int i = 0; i < nums.length; i++) {
int val = nums[i];
nums[i] = 0;
update(i, val);
}
}
public void update(int i, int val) {
// index + last '1' bit
int diff = val - nums[i];
nums[i] = val;
i++;
while (i < bit.length) {
bit[i] += diff;
i += i & (-i);
}
}
private int getSum(int i) {
// index - last '1' bit
i++;
int res = 0;
while (i > 0) {
res += bit[i];
i -= i & (-i);
}
return res;
}
public int sumRange(int i, int j) {
return getSum(j) - getSum(i-1);
}
}
/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* obj.update(i,val);
* int param_2 = obj.sumRange(i,j);
*/https://leetcode.com/problems/reverse-pairs/description/ https://leetcode.com/problems/count-of-smaller-numbers-after-self https://leetcode.com/problems/count-of-range-sum
Binary Search Tree
in-order asecending https://leetcode.com/problems/validate-binary-search-tree/description/ https://leetcode.com/problems/kth-smallest-element-in-a-bst/
Divide & Conquer
get result from children, then do something
https://leetcode.com/problems/longest-univalue-path/description/ https://leetcode.com/problems/count-univalue-subtrees/description/ https://leetcode.com/problems/subtree-of-another-tree/description/
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