05 August 2008
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:
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9
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:
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12
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.
这是一道树的题目,一般使用递归来做,主要就是考虑递归条件和结束条件。这道题思路还是比较明确的,目标是把从根到叶子节点的所有路径得到的整数都累加起来,递归条件即是把当前的sum乘以10并且加上当前节点传入下一函数,进行递归,最终把左右子树的总和相加。结束条件的话就是如果一个节点是叶子,那么我们应该累加到结果总和中,如果节点到了空节点,则不是叶子节点,不需要加入到结果中,直接返回0即可。算法的本质是一次先序遍历,所以时间是O(n),空间是栈大小,O(logn)。
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
int sum;
public int sumNumbers(TreeNode root) {
sum = 0;
dfs(root, 0);
return sum;
}
private void dfs(TreeNode node, int tempSum) {
if (node == null) {
return;
}
tempSum += node.val;
if (node.left == null && node.right == null) {
sum += tempSum;
}
tempSum *= 10;
dfs(node.left, tempSum);
dfs(node.right, tempSum);
}
}
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class Solution {
public int sumNumbers(TreeNode root) {
return dfs(root, 0);
}
private int dfs(TreeNode root, int sum) {
if (root == null) {
return 0;
}
if (root.left == null && root.right == null) {
return sum * 10 + root.val;
}
return dfs(root.left, sum * 10 + root.val) + dfs(root.right, sum * 10 + root.val);
}
}