Leetcode 257 - Binary tree paths

Note

  • Know how to check if a node is a leaf node
  • Use spread operator to pass by value (Otherwise changing an element of the array would propagate back to the original array)

Given the root of a binary tree, return all root-to-leaf paths in any order.

A leaf is a node with no children.

Example

img

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Input: root = [1,2,3,null,5]
Output: ["1->2->5","1->3"]
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/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {string[]}
*/
var binaryTreePaths = function(root) {
let result = [];
traverse(root, []);
result = result.map(arr => arr.join('->'));
return result;
function traverse(node, path) {
if (!node.left && !node.right) {
path.push(node.val);
result.push([...path]);
return;
}
path.push(node.val);
node.left && traverse(node.left, [...path]);
node.right && traverse(node.right, [...path]);
}
};

Backtracking

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/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {string[]}
*/
var binaryTreePaths = function(root) {
let result = [];
backtracking(root, []);
result = result.map(arr => arr.join('->'));
return result;

function backtracking(node, path) {
if (!node.left && !node.right) {
path.push(node.val);
result.push([...path]);
return;
}
path.push(node.val);
if (node.left) {
backtracking(node.left, path);
/* BACKTRACKING */
path.pop();
}
if (node.right) {
backtracking(node.right, path);
/* BACKTRACKING */
path.pop();
}
}
};