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1. Add leetcode solutions with tag amazon.
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leetcode/314. Binary Tree Vertical Order Traversal.md

Lines changed: 47 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -163,4 +163,50 @@ Output:
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return result;
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}
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}
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```
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```
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### Amazon session
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* Method 1: dfs + TreeMap
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```Java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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class Solution {
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// key: colNum, value: (key: rowNum, value: node values)
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private TreeMap<Integer, TreeMap<Integer, List<Integer>>> map;
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public List<List<Integer>> verticalOrder(TreeNode root) {
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this.map = new TreeMap<>();
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List<List<Integer>> result = new ArrayList<>();
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if(root == null) return result;
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// use recursion to add nodes to tree map.
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dfs(0, 0, root);
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// add to result;
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for(TreeMap<Integer, List<Integer>> out: map.values()){
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List<Integer> list = new ArrayList<>();
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for(List<Integer> value: out.values()){
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list.addAll(value);
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}
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result.add(list);
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}
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return result;
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}
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private void dfs(int x, int y, TreeNode node){
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if(node == null) return;
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else{
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TreeMap<Integer, List<Integer>> cur = map.getOrDefault(y, new TreeMap<>());
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List<Integer> list = cur.getOrDefault(x, new ArrayList<>());
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list.add(node.val);
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cur.put(x, list);
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map.put(y, cur);
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dfs(x + 1, y - 1, node.left);
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dfs(x + 1, y + 1, node.right);
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}
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}
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}
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```

leetcode/545. Boundary of Binary Tree.md

Lines changed: 46 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -104,4 +104,50 @@ So order them in anti-clockwise without duplicate nodes we have [1,2,4,7,8,9,10,
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if(node.right != null) dfs(node.right, list);
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}
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}
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```
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### Amazon Session
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* Method 1: DFS
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```Java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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class Solution {
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List<Integer> result;
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public List<Integer> boundaryOfBinaryTree(TreeNode root) {
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this.result = new ArrayList<>();
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if(root == null) return this.result;
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result.add(root.val);
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dfsLeft(root.left);
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dfsLeaves(root.left);
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dfsLeaves(root.right);
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dfsRight(root.right);
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return result;
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}
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private void dfsLeft(TreeNode node){
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if(node == null || (node.left == null && node.right == null)) return;
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result.add(node.val);
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dfsLeft(node.left == null ? node.right: node.left);
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}
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private void dfsRight(TreeNode node){
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if(node == null || (node.left == null && node.right == null)) return;
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dfsRight(node.right == null ? node.left: node.right);
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result.add(node.val);
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}
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private void dfsLeaves(TreeNode node){
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if(node == null) return;
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else if(node.left == null && node.right == null){
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this.result.add(node.val);
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return;
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}
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dfsLeaves(node.left);
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dfsLeaves(node.right);
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}
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}
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```

leetcode/863. All Nodes Distance K in Binary Tree.md

Lines changed: 68 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -88,4 +88,71 @@ Note:
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graph.put(node, neighbours);
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}
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}
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```
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```
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### Amazon session
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* Method 1: dfs + bfs
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```Java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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class Solution {
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private Map<TreeNode, List<TreeNode>> g;
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public List<Integer> distanceK(TreeNode root, TreeNode target, int K) {
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g = new HashMap<>();
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List<Integer> result = new ArrayList<>();
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dfs(root); // create a graph
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Queue<TreeNode> q = new LinkedList<>();
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q.offer(target);
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int level = 0;
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Set<TreeNode> visited = new HashSet<>();
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visited.add(target);
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while(!q.isEmpty() && level < K){
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int size = q.size();
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for(int sz = 0; sz < size; sz++){
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TreeNode cur = q.poll();
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if(g.containsKey(cur)){
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List<TreeNode> neighbours = g.get(cur);
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for(TreeNode n: neighbours){
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if(!visited.contains(n)){
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visited.add(n);
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q.offer(n);
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}
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}
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}
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}
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level++;
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}
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if(level == K){
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for(TreeNode n: q){
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result.add(n.val);
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}
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}
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return result;
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}
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private void dfs(TreeNode node){
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List<TreeNode> neighbours = g.getOrDefault(node, new ArrayList<>());
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if(node.left != null){
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neighbours.add(node.left);
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List<TreeNode> lefts = g.getOrDefault(node.left, new ArrayList<>());
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lefts.add(node);
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g.put(node.left, lefts);
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dfs(node.left);
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}
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if(node.right != null){
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neighbours.add(node.right);
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List<TreeNode> rights = g.getOrDefault(node.right, new ArrayList<>());
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rights.add(node);
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g.put(node.right, rights);
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dfs(node.right);
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}
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g.put(node, neighbours);
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}
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}
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```

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