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.
递归太好用了,over
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
def rootsum(root,value,res):
if root is None:
return res
if root.left is None and root.right is None:
res += value*10+root.val
else:
res = rootsum(root.left,value*10+root.val,res)
res = rootsum(root.right,value*10+root.val,res)
return res
class Solution(object):
def sumNumbers(self, root):
"""
:type root: TreeNode
:rtype: int
"""
res = 0
value = 0
res = rootsum(root,value,res)
return res
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
private int result = 0;
private void helper(TreeNode root, int path)
{
if(root == null)
return;
if(root.left == null && root.right == null)
{
path = path * 10 + root.val;
result += path;
}
if(root.left != null)
helper(root.left,path*10+root.val);
if(root.right != null)
helper(root.right, path*10+root.val);
}
public int sumNumbers(TreeNode root) {
int path = 0;
helper(root,path);
return result;
}
}