Portfolio

문제

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.

Constraints:

The number of nodes in the tree is in the range [0, 5000].

1000 <= Node.val <= 1000

1000 <= targetSum <= 1000

풀이

재영


/**
 * 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[][]}
 */
class Q {
    constructor() {
        this.arr = [];
        this.front = 0;
        this.rear = 0;
    }
    
    enqueue(val) {
        this.arr.push(val);
        this.rear += 1;
    }
    
    dequeue() {
        const value = this.arr[this.front];
       
        delete this.arr[this.front]
        this.front += 1;
        
        return value;
    }
    
    get length() {
        return this.rear - this.front    
    }
}

const nodes = ['left', 'right'];

const pathSum = (root, targetSum) => {
    const checkLeaf = (n) => !(n && nodes.some((direction) => n[direction]))
    
    if (checkLeaf(root)) { // 예외처리 1
        return root?.val === targetSum ? [[root?.val]] : []
    }
    
    const result = [];
    
    const queue = new Q();
    queue.enqueue([0, [], root])
    
    while (queue.length) {
        let [total, paths, now] = queue.dequeue();
        
        total += now.val;
        paths.push(now.val);
        
        if (checkLeaf(now)) {
            if (targetSum === total) {
                result.push(paths)
            }
        }
        
        nodes.forEach((direction) => {
            if (now[direction] !== null) {
                queue.enqueue([total, [...paths], now[direction]])
            }
        })
    }
    
    return result;
};

효성


var pathSum = function(root, targetSum) {
    if(!root) {
        return [];
    }
    
    const answer = [];

    const dfs = (node, sum, nodePath) => {
        if(!node.left && !node.right) {
            const total = sum + node.val;
            if(total === targetSum) {
                answer.push([...nodePath, node.val]);
            }
            return;
        }

        node.right && dfs(node.right, sum + node.val, [...nodePath, node.val]);
        node.left && dfs(node.left, sum + node.val, [...nodePath, node.val]);
    }

    dfs(root, 0, []);

    return answer;
};

메모리 좀 더 좋게 해봤어요.


var pathSum = function(root, targetSum) {
    const answer = [];

    if(!root) {
        return answer;
    }
    
    const dfs = (node, sum, nodePath) => {
        if(!node.left && !node.right) {
            const total = sum + node.val;
            if(total === targetSum) {
                answer.push([...nodePath, node.val]);
            }
            return;
        }

        nodePath.push(node.val);
        node.right && dfs(node.right, sum + node.val, nodePath);
        node.left && dfs(node.left, sum + node.val, nodePath);
        nodePath.pop();
    }

    dfs(root, 0, []);

    return answer;
};

Path Sum II

문제

풀이