Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path 1->2->3 which represents the number 123.
Find the total sum of all root-to-leaf numbers.
Note: A leaf is a node with no children.
Example:
Input: [1,2,3]
1
/ \
2 3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.
Example 2:
Input: [4,9,0,5,1]
4
/ \
9 0
/ \
5 1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.
Solution：DFS， Recursion -- Top- down
- Top-down扔役，每次都把
parent的值 * 10 + current value朝捆，就是當(dāng)前節(jié)點(diǎn)的value - 迭代函數(shù)維護(hù)2個變量：
int parentSum, int[] result - Base Case：
if (root == null) {
return;
}
if (root.left == null && root.right == null) {
int currentNum = parentSum * 10 + root.val;
result[0] += currentNum;
return;
}
- 將
ParentSum * 10 + current Value傳入左右子節(jié)點(diǎn)的遞歸函數(shù)中繼續(xù)求解
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int sumNumbers(TreeNode root) {
if (root == null) {
return 0;
}
int[] result = { 0 };
sumNumbers (root, 0, result);
return result[0];
}
private void sumNumbers (TreeNode root, int parentSum, int[] result) {
if (root == null) {
return;
}
if (root.left == null && root.right == null) {
int currentNum = parentSum * 10 + root.val;
result[0] += currentNum;
return;
}
sumNumbers (root.left, parentSum * 10 + root.val, result);
sumNumbers (root.right, parentSum * 10 + root.val, result);
}
}
