129. sum root to leaf numbers

A binary tree question.

The general ieda is using preorder traversal in recursion way. Iterative solution is also accepted.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
class  {
int sum;
public int sumNumbers(TreeNode root) {
sum = 0;
dfs(root, 0);
return sum;
}

public void dfs(TreeNode node, int cur) {
if (node == null) return;
cur = cur * 10 + node.val;
if (node.left == null && node.right == null) {
sum += cur;
return;
}
if (node.left != null) dfs(node.left, cur);
if (node.right != null) dfs(node.right, cur);
}

public int iterative(TreeNode root) {
if (root == null) return 0;
Stack<TreeNode> stack = new Stack<>();
Stack<Integer> sum = new Stack<>();
stack.push(root);
sum.push(root.val);
int res = 0;
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
int cur = sum.pop();
if (node.left == null && node.right == null) {
res += cur;
}
if (node.right != null) {
stack.push(node.right);
sum.push(cur * 10 + node.right.val);
}
if (node.left != null) {
stack.push(node.left);
sum.push(cur * 10 + node.left.val);
}
}
return res;
}
}