/**
* 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)
}
leetcode
solution
Here is the link for the problem:
https://leetcode.com/problems/sum-root-to-leaf-numbers/
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