2014
11-19

LeetCode-Sum Root to Leaf Numbers[二叉树]

Sum Root to Leaf Numbers

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.

// LeetCode, Decode Ways
// 时间复杂度O(n)，空间复杂度O(logn)
class Solution {
public:
int sumNumbers(TreeNode *root) {
return dfs(root, 0);
}
private:
int dfs(TreeNode *root, int sum) {
if (root == nullptr) return 0;
if (root->left == nullptr && root->right == nullptr)
return sum * 10 + root->val;

return dfs(root->left, sum * 10 + root->val) +
dfs(root->right, sum * 10 + root->val);
}
};


Java代码:

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:

def sumNumRecr(self,root, parent):
if root.left == None and root.right == None:
self.sum +=(parent*10 + root.val)

if root.left != None:
self.sumNumRecr(root.left, parent*10 + root.val)

if root.right != None:
self.sumNumRecr( root.right, parent*10 + root.val)

# @param root, a tree node
# @return an integer
def sumNumbers(self, root):
if(root == None):
return 0
self.sum = 0
self.sumNumRecr(root,0)
return self.sum