(Leet Code JS)Path Sum II

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].
• -1000 <= Node.val <= 1000
• -1000 <= targetSum <= 1000

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;
};