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Commit 5e2e374

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feat: add solutions to lc problem: No.1483
No.1483.Kth Ancestor of a Tree Node
1 parent d8b21c0 commit 5e2e374

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3 files changed

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‎solution/1400-1499/1483.Kth Ancestor of a Tree Node/README.md‎

Lines changed: 45 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -57,9 +57,9 @@ treeAncestor.getKthAncestor(6, 3); // 返回 -1 因为不存在满足要求的
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**方法一:动态规划 + 倍增**
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题目要我们寻找节点 `node` 的第 $k$ 个祖先节点,如果暴力求解,需要从 `node` 开始向上遍历 $k$ 次,时间复杂度为 $O(k),ドル显然会超时。
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题目要我们寻找节点 $node$ 的第 $k$ 个祖先节点,如果暴力求解,需要从 $node$ 开始向上遍历 $k$ 次,时间复杂度为 $O(k),ドル显然会超时。
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我们可以使用动态规划的思想,结合倍增的思想来处理。
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我们可以使用动态规划,结合倍增的思想来处理。
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我们定义 $p[i][j]$ 表示节点 $i$ 的第 2ドル^j$ 个祖先节点,即 $p[i][j]$ 表示节点 $i$ 向上走 2ドル^j$ 步的节点。那么我们可以得到状态转移方程:
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@@ -71,7 +71,7 @@ $$
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之后对于每次查询,我们可以把 $k$ 拆成二进制的表示形式,然后根据二进制中 1ドル$ 的位置,累计向上查询,最终得到节点 $node$ 的第 $k$ 个祖先节点。
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时间复杂度:初始化为 $O(n \times \log n),ドル查询为 $O(\log n)$。空间复杂度:$O(n \times \log n)$。其中 $n$ 为树的节点数。
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时间复杂度方面,初始化为 $O(n \times \log n),ドル查询为 $O(\log n)$。空间复杂度$O(n \times \log n)$。其中 $n$ 为树的节点数。
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<!-- tabs:start -->
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@@ -240,6 +240,48 @@ func (this *TreeAncestor) GetKthAncestor(node int, k int) int {
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*/
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```
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### **TypeScript**
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```ts
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class TreeAncestor {
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private p: number[][];
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constructor(n: number, parent: number[]) {
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const p = new Array(n).fill(0).map(() => new Array(18).fill(-1));
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for (let i = 0; i < n; ++i) {
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p[i][0] = parent[i];
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}
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for (let i = 0; i < n; ++i) {
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for (let j = 1; j < 18; ++j) {
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if (p[i][j - 1] === -1) {
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continue;
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}
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p[i][j] = p[p[i][j - 1]][j - 1];
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}
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}
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this.p = p;
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}
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getKthAncestor(node: number, k: number): number {
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for (let i = 17; i >= 0; --i) {
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if (((k >> i) & 1) === 1) {
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node = this.p[node][i];
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if (node === -1) {
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break;
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}
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}
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}
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return node;
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}
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}
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/**
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* Your TreeAncestor object will be instantiated and called as such:
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* var obj = new TreeAncestor(n, parent)
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* var param_1 = obj.getKthAncestor(node,k)
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*/
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```
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### **...**
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```

‎solution/1400-1499/1483.Kth Ancestor of a Tree Node/README_EN.md‎

Lines changed: 42 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -208,6 +208,48 @@ func (this *TreeAncestor) GetKthAncestor(node int, k int) int {
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*/
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```
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### **TypeScript**
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```ts
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class TreeAncestor {
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private p: number[][];
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constructor(n: number, parent: number[]) {
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const p = new Array(n).fill(0).map(() => new Array(18).fill(-1));
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for (let i = 0; i < n; ++i) {
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p[i][0] = parent[i];
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}
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for (let i = 0; i < n; ++i) {
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for (let j = 1; j < 18; ++j) {
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if (p[i][j - 1] === -1) {
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continue;
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}
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p[i][j] = p[p[i][j - 1]][j - 1];
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}
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}
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this.p = p;
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}
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getKthAncestor(node: number, k: number): number {
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for (let i = 17; i >= 0; --i) {
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if (((k >> i) & 1) === 1) {
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node = this.p[node][i];
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if (node === -1) {
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break;
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}
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}
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}
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return node;
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}
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}
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/**
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* Your TreeAncestor object will be instantiated and called as such:
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* var obj = new TreeAncestor(n, parent)
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* var param_1 = obj.getKthAncestor(node,k)
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*/
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```
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### **...**
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```
Lines changed: 37 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,37 @@
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class TreeAncestor {
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private p: number[][];
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constructor(n: number, parent: number[]) {
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const p = new Array(n).fill(0).map(() => new Array(18).fill(-1));
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for (let i = 0; i < n; ++i) {
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p[i][0] = parent[i];
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}
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for (let i = 0; i < n; ++i) {
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for (let j = 1; j < 18; ++j) {
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if (p[i][j - 1] === -1) {
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continue;
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}
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p[i][j] = p[p[i][j - 1]][j - 1];
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}
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}
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this.p = p;
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}
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getKthAncestor(node: number, k: number): number {
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for (let i = 17; i >= 0; --i) {
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if (((k >> i) & 1) === 1) {
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node = this.p[node][i];
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if (node === -1) {
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break;
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}
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}
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}
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return node;
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}
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}
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/**
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* Your TreeAncestor object will be instantiated and called as such:
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* var obj = new TreeAncestor(n, parent)
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* var param_1 = obj.getKthAncestor(node,k)
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*/

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