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(Leet Code JS)Path Sum II
(Leet Code JS)Path Sum II
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113. Path Sum II
Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equals targetSum. Each path should be returned as a list of the node values, not node references. A root-to-leaf path is a path starting from the root and ending at any leaf node. A leaf is a node with no children. Example 1: Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22 Output: [[5,4,11,2],[5,8,4,5]] Explanation: There are two paths whose sum equals targetSum: 5 + 4 + 11 + 2 = 22 5 + 8 + 4 + 5 = 22 Example 2: Input: root = [1,2,3], targetSum = 5 Output: [] Example 3: Input: root = [1,2], targetSum = 0 Output: [] Constraints: The number of nodes in the tree is in the range [0, 5000].
[0, 5000]. -1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000 728x90
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리프 노드이면서 해당 노드까지의 합이 targetSum과 같은 경로를 출력.
dfs를 통해 진행.
dfs 함수
우선 현재까지의 sum이 targetSum과 같은지, 현재 노드의 left와 rigth가 모두 null 인 경우 answer에 현재까지의 경로를 push 후 리턴.
그렇지 않다면 왼쪽과 오른쪽 노드를 dfs 탐색 진행하면 된다.
/** * 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 * @param {number} targetSum * @return {number[][]} */ const dfs = (node, sumArray, sum, answer, targetSum) => { if(sum === targetSum && node.left === null && node.right === null){ answer.push(sumArray); return; } if(node.left !== null){ dfs(node.left, [...sumArray, node.left.val], sum + node.left.val, answer, targetSum); } if(node.right !== null){ dfs(node.right, [...sumArray, node.right.val], sum + node.right.val, answer, targetSum); } }; const pathSum = (root, targetSum) => { const answer = []; if(root !== null){ dfs(root, [root.val], root.val, answer, targetSum); } return answer; };
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from http://eunchanee.tistory.com/611 by ccl(A) rewrite - 2021-11-15 01:26:35