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.
For example,
1
/ \
2 3
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Return the sum = 12 + 13 = 25.
一刷
題解：DFS。leaf的左右子節(jié)點均為空。
Time Complexity - O(n)父叙， Space Complexity - O(n)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int sumNumbers(TreeNode root) {
return sum(root, 0);
}
public int sum(TreeNode n, int s){
if(n == null) return 0;
if(n.right == null && n.left == null) return s*10 + n.val;
return sum(n.left, s*10 + n.val) + sum(n.right, s*10 + n.val);
}
}
二刷：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
int total = 0;
public int sumNumbers(TreeNode root) {
if(root == null) return 0;
sum(0, root);
return total;
}
public void sum(int prev, TreeNode root){
if(root == null) return;
if(root.left == null && root.right == null) {
total += prev*10 + root.val;
return;
}
sum(prev*10 + root.val, root.left);
sum(prev*10 + root.val, root.right);
}
}
三刷
注意要判斷葉子，否則[1,2]這種結(jié)構(gòu)會返回1+12
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int sumNumbers(TreeNode root) {
return sumNumbers(root, 0);
}
public int sumNumbers(TreeNode root, int sum){
if(root == null) return 0;
if(root.left == null && root.right == null) return sum*10 + root.val;
return sumNumbers(root.left, 10*sum + root.val) + sumNumbers(root.right, 10*sum + root.val);
}
}
