problem: 0863 all nodes distance k in binary tree

Author: cychiu8

0863 All Nodes Distance K in Binary Tree/0863 All Nodes Distance K in Binary Tree.md

@@ -0,0 +1,79 @@
+# 0863. All Nodes Distance K in Binary Tree
+
+* Difficulty: medium
+* Link: https://leetcode.com/problems/all-nodes-distance-k-in-binary-tree/
+* Topics: DFS-BFS
+
+# Clarification
+
+1. Check the inputs and outputs
+    - INPUT: TreeNode, TreeNode, k
+    - OUTPUT: List[int]
+
+# Naive Solution
+
+### Thought Process
+
+1. transfer the tree to graph
+2. start from the target node from the graph using bfs
+3. return the level k
+- Implement
+    
+    ```python
+    # Definition for a binary tree node.
+    # class TreeNode:
+    #     def __init__(self, x):
+    #         self.val = x
+    #         self.left = None
+    #         self.right = None
+    
+    class Solution:
+        def distanceK(self, root: TreeNode, target: TreeNode, k: int) -> List[int]:
+            
+            if not root:
+                return []
+            
+            #1. transfer the tree to graph
+            graph = {}
+            q = collections.deque()
+            q.append(root)
+            
+            while q:
+                current = q.popleft()
+                if not graph.get(current.val):
+                    graph[current.val] = []
+                if current.left:
+                    leftVal = current.left.val
+                    graph[current.val].append(leftVal)
+                    graph[leftVal] = [current.val]
+                    q.append(current.left)
+                if current.right:
+                    rightVal = current.right.val
+                    graph[current.val].append(rightVal)
+                    graph[rightVal] = [current.val]
+                    q.append(current.right)
+            #2. start from the target node from the graph using bfs
+            q.append(target.val)
+            level = 0
+            result = []
+            while level <= k:
+                if not q:
+                    return []
+                lenq = len(q)
+                result = []
+                for i in range(lenq):
+                    current = q.popleft()
+                    result.append(current)
+                    for n in graph.get(current):
+                        graph[n].remove(current)
+                        q.append(n)
+                level += 1
+            #3. return the level k
+            return result
+    ```
+    
+
+### Complexity
+
+- Time complexity:O(N)
+- Space complexity:O(N)