/** * 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 {number} */ var sumNumbers = function(root) { if (!root) return 0; var total = []; helper(root, 0, total); var sum = 0; for (var i = 0; i < total.length; i++) { sum += total[i]; } return sum; }; var helper = function(root, sum, total) { sum = 10 * sum + root.val; if (root.left === null && root.right === null) { total.push(sum); return; } if (root.left) { helper(root.left, sum, total); } if (root.right) { helper(root.right, sum, total); } }; // a better and more concise solution var sumNumbers = function(root) { return helper(root, 0); }; var helper = function(root, sum) { if (!root) return 0; if (root.left === null && root.right === null) { return 10 * sum + root.val; } return helper(root.left, 10 * sum + root.val) + helper(root.right, 10 * sum + root.val); }; // a third DFS method. Top-down var sumNumbers = function(root) { if (!root) return 0; return helper(root, 0, 0); }; function helper(root, currNum, sum) { if (!root) return sum; currNum = root.val + 10 * currNum; if (!root.left && !root.right) { sum += currNum; return sum; } return helper(root.left, currNum, sum) + helper(root.right, currNum, sum) }