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.
给一个只含有数字0-9的二叉树，每一个从根节点到叶节点的路径代表一个数字，求所有这些数字的和。
解法1：递归，累加所有路径上节点的值，每多一层之前的值要扩大十倍。
解法2: 迭代，while循环，用stack或queue来存下一层的节点和之前的和。
Java:
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);
}
Python:
# Time: O(n)
# Space: O(h), h is height of binary tree
class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution(object):
# @param root, a tree node
# @return an integer
def sumNumbers(self, root):
return self.sumNumbersRecu(root, 0)
def sumNumbersRecu(self, root, num):
if root is None:
return 0
if root.left is None and root.right is None:
return num * 10 + root.val
return self.sumNumbersRecu(root.left, num * 10 + root.val) + self.sumNumbersRecu(root.right, num * 10 + root.val)
Python: bfs + stack
def sumNumbers1(self, root):
if not root:
return 0
stack, res = [(root, root.val)], 0
while stack:
node, value = stack.pop()
if node:
if not node.left and not node.right:
res += value
if node.right:
stack.append((node.right, value*10+node.right.val))
if node.left:
stack.append((node.left, value*10+node.left.val))
return res
Python: bfs + queue
#
def sumNumbers2(self, root):
if not root:
return 0
queue, res = collections.deque([(root, root.val)]), 0
while queue:
node, value = queue.popleft()
if node:
if not node.left and not node.right:
res += value
if node.left:
queue.append((node.left, value*10+node.left.val))
if node.right:
queue.append((node.right, value*10+node.right.val))
return res
Python: Recursive
def sumNumbers(self, root):
self.res = 0
self.dfs(root, 0)
return self.res
def dfs(self, root, value):
if root:
#if not root.left and not root.right:
# self.res += value*10 + root.val
self.dfs(root.left, value*10+root.val)
#if not root.left and not root.right:
# self.res += value*10 + root.val
self.dfs(root.right, value*10+root.val)
if not root.left and not root.right:
self.res += value*10 + root.val
C++:
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int sumNumbers(TreeNode *root) {
return sumNumbersDFS(root, 0);
}
int sumNumbersDFS(TreeNode *root, int sum) {
if (!root) return 0;
sum = sum * 10 + root->val;
if (!root->left && !root->right) return sum;
return sumNumbersDFS(root->left, sum) + sumNumbersDFS(root->right, sum);
}
};
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