feat: add ts solution to problem: No.863 (doocs#2913)

Author: rain84

solution/0800-0899/0863.All Nodes Distance K in Binary Tree/README.md

@@ -302,6 +302,64 @@ class Solution:
         return ans
 ```
 
+#### TypeScript
+
+```ts
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *     val: number
+ *     left: TreeNode | null
+ *     right: TreeNode | null
+ *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.left = (left===undefined ? null : left)
+ *         this.right = (right===undefined ? null : right)
+ *     }
+ * }
+ */
+
+function distanceK(root: TreeNode | null, target: TreeNode | null, k: number): number[] {
+    if (!root) return [0];
+
+    const g: Record<number, number[]> = {};
+
+    const dfs = (node: TreeNode | null, parent: TreeNode | null = null) => {
+        if (!node) return;
+
+        g[node.val] ??= [];
+        if (parent) g[node.val].push(parent.val);
+        if (node.left) g[node.val].push(node.left.val);
+        if (node.right) g[node.val].push(node.right.val);
+
+        dfs(node.left, node);
+        dfs(node.right, node);
+    };
+
+    dfs(root);
+
+    const vis = new Set<number>();
+    let q = [target!.val];
+
+    while (q.length) {
+        if (!k--) return q;
+
+        const nextQ: number[] = [];
+
+        for (const x of q) {
+            if (vis.has(x)) continue;
+
+            vis.add(x);
+            nextQ.push(...g[x].filter(x => !vis.has(x)));
+        }
+
+        q = nextQ;
+    }
+
+    return [];
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/0800-0899/0863.All Nodes Distance K in Binary Tree/README_EN.md

@@ -242,6 +242,64 @@ func distanceK(root *TreeNode, target *TreeNode, k int) []int {
 }
 ```
 
+#### TypeScript
+
+```ts
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *     val: number
+ *     left: TreeNode | null
+ *     right: TreeNode | null
+ *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.left = (left===undefined ? null : left)
+ *         this.right = (right===undefined ? null : right)
+ *     }
+ * }
+ */
+
+function distanceK(root: TreeNode | null, target: TreeNode | null, k: number): number[] {
+    if (!root) return [0];
+
+    const g: Record<number, number[]> = {};
+
+    const dfs = (node: TreeNode | null, parent: TreeNode | null = null) => {
+        if (!node) return;
+
+        g[node.val] ??= [];
+        if (parent) g[node.val].push(parent.val);
+        if (node.left) g[node.val].push(node.left.val);
+        if (node.right) g[node.val].push(node.right.val);
+
+        dfs(node.left, node);
+        dfs(node.right, node);
+    };
+
+    dfs(root);
+
+    const vis = new Set<number>();
+    let q = [target!.val];
+
+    while (q.length) {
+        if (!k--) return q;
+
+        const nextQ: number[] = [];
+
+        for (const x of q) {
+            if (vis.has(x)) continue;
+
+            vis.add(x);
+            nextQ.push(...g[x].filter(x => !vis.has(x)));
+        }
+
+        q = nextQ;
+    }
+
+    return [];
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/0800-0899/0863.All Nodes Distance K in Binary Tree/Solution.ts

@@ -0,0 +1,53 @@
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *     val: number
+ *     left: TreeNode | null
+ *     right: TreeNode | null
+ *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.left = (left===undefined ? null : left)
+ *         this.right = (right===undefined ? null : right)
+ *     }
+ * }
+ */
+
+function distanceK(root: TreeNode | null, target: TreeNode | null, k: number): number[] {
+    if (!root) return [0];
+
+    const g: Record<number, number[]> = {};
+
+    const dfs = (node: TreeNode | null, parent: TreeNode | null = null) => {
+        if (!node) return;
+
+        g[node.val] ??= [];
+        if (parent) g[node.val].push(parent.val);
+        if (node.left) g[node.val].push(node.left.val);
+        if (node.right) g[node.val].push(node.right.val);
+
+        dfs(node.left, node);
+        dfs(node.right, node);
+    };
+
+    dfs(root);
+
+    const vis = new Set<number>();
+    let q = [target!.val];
+
+    while (q.length) {
+        if (!k--) return q;
+
+        const nextQ: number[] = [];
+
+        for (const x of q) {
+            if (vis.has(x)) continue;
+
+            vis.add(x);
+            nextQ.push(...g[x].filter(x => !vis.has(x)));
+        }
+
+        q = nextQ;
+    }
+
+    return [];
+}