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社区首页 >专栏 >Leetcode: Sum Root to Leaf Numbers

Leetcode: Sum Root to Leaf Numbers

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卡尔曼和玻尔兹曼谁曼
发布2019-01-22 16:00:53
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发布2019-01-22 16:00:53
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题目:

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.

思路分析:

求二叉树根到叶节点的路径和。

采用二叉树的深度优先遍历,遍历过程中记录路径并求和。

C++参考代码:

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution
{
private:
    //二叉树深度优先遍历
    void dfs(TreeNode *root, int &sum, int num)
    {
    	if (!root->left && !root->right) sum += num;
    	if (root->left) dfs(root->left, sum, num * 10 + root->left->val);
    	if (root->right) dfs(root->right, sum, num * 10 + root->right->val);
    }
    
public:
    int sumNumbers(TreeNode *root)
    {
    	if (!root) return 0;
    	int sum = 0;
    	dfs(root, sum, root->val);
    	return sum;
    }
};

C#参考代码:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution
{
    private int sum = 0;
    private void DFS(TreeNode root, int num)
    {
        if (root.left == null && root.right == null) sum += num;
        if (root.left != null) DFS(root.left, num * 10 + root.left.val);
        if (root.right != null) DFS(root.right, num * 10 + root.right.val);
    }

    public int SumNumbers(TreeNode root)
    {
        if (root == null) return 0;
        DFS(root, root.val);
        return sum;
    }
}
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原始发表:2015年04月07日,如有侵权请联系 cloudcommunity@tencent.com 删除

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