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253. Lowest Common Ancestor of a Binary Search Tree

· 2 min read
Shi Xinyu
Front End Developer

Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Example

Example 1:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 Output: 6 Explanation: The LCA of nodes 2 and 8 is 6.

Example 2:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4 Output: 2 Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/

/**
* @param {TreeNode} root
* @param {TreeNode} p
* @param {TreeNode} q
* @return {TreeNode}
*/
var lowestCommonAncestor = function (root, p, q) {
let pPath = [];
let qPath = [];
let path = [];
function travel(node) {
if (node) {
path.push(node);
if (node === p) {
pPath = path.slice(0);
}
if (node === q) {
qPath = path.slice(0);
}
travel(node.left);
travel(node.right);
path.pop();
}
}
travel(root);

let i = 0;
for (; i < pPath.length || i < qPath.length; i++) {
if (pPath[i] !== qPath[i]) {
break;
}
}
return pPath[i - 1];
};

867. Transpose Matrix

· One min read
Shi Xinyu
Front End Developer

Given a 2D integer array matrix, return the transpose of matrix.

The transpose of a matrix is the matrix flipped over its main diagonal, switching the matrix's row and column indices. link: https://leetcode.com/problems/transpose-matrix/

Example

Example 1:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [[1,4,7],[2,5,8],[3,6,9]]

Example 2:

Input: matrix = [[1,2,3],[4,5,6]] Output: [[1,4],[2,5],[3,6]]

/**
* @param {number[][]} matrix
* @return {number[][]}
*/
var transpose = function (matrix) {
if (!matrix.length) return;
const numOfCols = matrix[0].length;
const numOfRows = matrix.length;
const result = [];
for (let i = 0; i < numOfCols; i++) {
result.push([]);
}
for (let i = 0; i < numOfCols; i++) {
for (let j = 0; j < numOfRows; j++) {
result[i][j] = matrix[j][i];
}
}
return result;
};

About Me

· One min read
Shi Xinyu
Front End Developer

I am Shi, author of this blog. I am a software engineer and I love to write about technology. Hope you enjoy my blog.