about Binary Tree
Tue, Feb 16, 2021
2-minute read
learning binary tree
Use java to realize a simple Customer Account project.
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
Queue<TreeNode> q = new ArrayDeque<>();
if (root != null) {
q.add(root);
}
while(!q.isEmpty()) {
int size = q.size();
List<Integer> list = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode node = q.poll();
list.add(node.val);
if (node.left != null) {
q.add(node.left);
}
if (node.right != null) {
q.add(node.right);
}
}
res.add(list);
}
return res;
}
}
BFS: level order
BFS: shortest path
also a grid search
DFS: grid struction: island
1. DFS basic structure
void traverse(TreeNode root) {
// base case
if (root == null) {
return;
}
//
traverse(root.left);
traverse(root.right);
}
grid (r, c) , its neigbor: (r -1, c), (r + 1, c), (r, c - 1), (r, c + 1)
2. DFS grid basic structure
void dfs(int[][] grid, int r, int c) {
// base case, (r, c)exceed the board
if (!inArea(grid, r, c)) {
return;
}
//
dfs(grid, r -1, c);
dfs(grid, r + 1, c);
dfs(grid, r, c - 1);
dfs(grid, r, c + 1);
}
boolean inArea(int[][] grid, int r, int c) {
return 0 <= r && r < grid.length &&
&& 0 <= c && c < grid[0].length;
}
3. record the visited grid
- 0 –occean
- 1 –island (unvisited)
- 2 –island (visited)
void dfs(int[][] grid, int r, int c) {
// base case, (r, c)exceed the board
if (!inArea(grid, r, c)) {
return;
}
if (grid[r][c] != 1) {
return;
}
grid[r][c] = 2
//
dfs(grid, r -1, c);
dfs(grid, r + 1, c);
dfs(grid, r, c - 1);
dfs(grid, r, c + 1);
}
boolean inArea(int[][] grid, int r, int c) {
return 0 <= r && r < grid.length &&
&& 0 <= c && c < grid[0].length;
}
LeetCode Max Area of Island LeetCode Making A Large Island LeetCode Island Perimeter