國產印表機納思達生產基地落戶合肥:總投資 70 億元,建成後年產量約 200 萬臺
阿新 • • 發佈:2021-12-06
題目連結:
給你二叉樹的根節點 root
和一個整數目標和 targetSum
,找出所有 從根節點到葉子節點 路徑總和等於給定目標和的路徑。
葉子節點
示例 1:
輸入:root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
輸出:[[5,4,11,2],[5,8,4,5]]
示例 2:
輸入:root = [1,2,3], targetSum = 5
輸出:[]
示例 3:
輸入:root = [1,2], targetSum = 0
輸出:[]
解題思路
該題與
遞迴法
程式碼(C++)
//遞迴 class Solution { public: void traversal(TreeNode* node, vector<int>& path, vector<vector<int>>& result, int count) { if(!node->left && !node->right) { if (count == 0) result.push_back(path);return; } if (node->left) { path.push_back(node->left->val); count -= node->left->val; traversal(node->left, path, result, count); //遞迴 count += node->left->val; //回溯 path.pop_back(); //回溯 }if (node->right) { path.push_back(node->right->val); count -= node->right->val; traversal(node->right, path, result, count); //遞迴 count += node->right->val; //回溯 path.pop_back(); //回溯 } } vector<vector<int>> pathSum(TreeNode* root, int targetSum) { vector<vector<int>> result; if (root == nullptr) return result; vector<int> path; path.push_back(root->val); traversal(root, path, result, targetSum - root->val); return result; } };
程式碼(JavaScript)
/** * @param {TreeNode} root * @param {number} targetSum * @return {number[][]} */ function traversal(node, path, result, count) { if (!node.left && !node.right) { if (count === 0) result.push([...path]); // 不能寫result.push(path), 要深拷貝 return; } if (node.left) { path.push(node.left.val); count -= node.left.val; traversal(node.left, path, result, count); //遞迴 count += node.left.val; //回溯 path.pop(); //回溯 } if (node.right) { path.push(node.right.val); count -= node.right.val; traversal(node.right, path, result, count); //遞迴 count += node.right.val; //回溯 path.pop(); //回溯 } return false; } var pathSum = function(root, targetSum) { let result = []; if (root === null) return result; let path = []; path.push(root.val); traversal(root, path, result, targetSum - root.val); return result; };