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Commit 2bfcb9f

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feat: add typescript solution to lc problem: No.0987 (doocs#625)
No.0987.Vertical Order Traversal of a Binary Tree
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‎solution/0900-0999/0987.Vertical Order Traversal of a Binary Tree/README.md‎

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```
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### **TypeScript**
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```ts
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/**
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* Definition for a binary tree node.
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* class TreeNode {
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* val: number
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* left: TreeNode | null
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* right: TreeNode | null
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* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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* }
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*/
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function verticalTraversal(root: TreeNode | null): number[][] {
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let solution = [];
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dfs(root, 0, 0, solution);
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// 优先依据i=2排序, 然后依据i=1排序
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solution.sort(compare);
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// console.log(solution);
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let ans = [];
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let pre = Number.MIN_SAFE_INTEGER;
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for (let node of solution) {
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const [val, , idx] = node;
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if (idx != pre) {
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ans.push([]);
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pre = idx;
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}
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ans[ans.length - 1].push(val);
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}
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return ans;
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};
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function compare(a: Array<number>, b: Array<number>) {
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const [a0, a1, a2] = a, [b0, b1, b2] = b;
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if (a2 == b2) {
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if (a1 == b1) {
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return a0 - b0;
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}
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return a1 - b1;
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}
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return a2 - b2;
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}
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function dfs(root: TreeNode | null, depth: number, idx: number, solution: Array<Array<number>>) {
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if (!root) return;
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solution.push([root.val, depth, idx]);
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dfs(root.left, depth + 1, idx - 1, solution);
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dfs(root.right, depth + 1, idx + 1, solution);
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}
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```
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### **...**
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```

‎solution/0900-0999/0987.Vertical Order Traversal of a Binary Tree/README_EN.md‎

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@@ -74,6 +74,61 @@ Note that the solution remains the same since 5 and 6 are in the same location a
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```
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### **TypeScript**
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```ts
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/**
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* Definition for a binary tree node.
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* class TreeNode {
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* val: number
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* left: TreeNode | null
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* right: TreeNode | null
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* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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* }
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*/
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function verticalTraversal(root: TreeNode | null): number[][] {
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let solution = [];
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dfs(root, 0, 0, solution);
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// 优先依据i=2排序, 然后依据i=1排序
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solution.sort(compare);
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// console.log(solution);
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let ans = [];
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let pre = Number.MIN_SAFE_INTEGER;
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for (let node of solution) {
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const [val, , idx] = node;
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if (idx != pre) {
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ans.push([]);
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pre = idx;
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}
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ans[ans.length - 1].push(val);
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}
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return ans;
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};
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function compare(a: Array<number>, b: Array<number>) {
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const [a0, a1, a2] = a, [b0, b1, b2] = b;
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if (a2 == b2) {
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if (a1 == b1) {
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return a0 - b0;
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}
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return a1 - b1;
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}
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return a2 - b2;
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}
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function dfs(root: TreeNode | null, depth: number, idx: number, solution: Array<Array<number>>) {
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if (!root) return;
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solution.push([root.val, depth, idx]);
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dfs(root.left, depth + 1, idx - 1, solution);
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dfs(root.right, depth + 1, idx + 1, solution);
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}
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```
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### **...**
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```
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/**
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* Definition for a binary tree node.
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* class TreeNode {
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* val: number
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* left: TreeNode | null
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* right: TreeNode | null
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* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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* }
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*/
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function verticalTraversal(root: TreeNode | null): number[][] {
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let solution = [];
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dfs(root, 0, 0, solution);
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// 优先依据i=2排序, 然后依据i=1排序
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solution.sort(compare);
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// console.log(solution);
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let ans = [];
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let pre = Number.MIN_SAFE_INTEGER;
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for (let node of solution) {
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const [val, , idx] = node;
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if (idx != pre) {
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ans.push([]);
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pre = idx;
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}
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ans[ans.length - 1].push(val);
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}
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return ans;
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};
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function compare(a: Array<number>, b: Array<number>) {
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const [a0, a1, a2] = a, [b0, b1, b2] = b;
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if (a2 == b2) {
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if (a1 == b1) {
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return a0 - b0;
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}
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return a1 - b1;
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}
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return a2 - b2;
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}
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function dfs(root: TreeNode | null, depth: number, idx: number, solution: Array<Array<number>>) {
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if (!root) return;
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solution.push([root.val, depth, idx]);
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dfs(root.left, depth + 1, idx - 1, solution);
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dfs(root.right, depth + 1, idx + 1, solution);
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}

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