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Create Solution.ts
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  • solution/2000-2099/2096.Step-By-Step Directions From a Binary Tree Node to Another

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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 getDirections(root: TreeNode | null, startValue: number, destValue: number): string {
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const lca = (node: TreeNode | null, p: number, q: number): TreeNode | null => {
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if (node === null || node.val === p || node.val === q) {
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return node;
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}
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const left = lca(node.left, p, q);
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const right = lca(node.right, p, q);
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if (left !== null && right !== null) {
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return node;
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}
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return left !== null ? left : right;
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};
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const dfs = (node: TreeNode | null, x: number, path: string[]): boolean => {
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if (node === null) {
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return false;
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}
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if (node.val === x) {
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return true;
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}
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path.push('L');
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if (dfs(node.left, x, path)) {
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return true;
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}
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path[path.length - 1] = 'R';
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if (dfs(node.right, x, path)) {
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return true;
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}
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path.pop();
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return false;
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};
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const node = lca(root, startValue, destValue);
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const pathToStart: string[] = [];
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const pathToDest: string[] = [];
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dfs(node, startValue, pathToStart);
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dfs(node, destValue, pathToDest);
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return 'U'.repeat(pathToStart.length) + pathToDest.join('');
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}

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