LeetCode_Sum Root to Leaf Numbers

一.题目

Sum Root to Leaf Numbers

  Total Accepted: 47437 Total Submissions: 156443My Submissions

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.

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二.解题技巧

    这道题只是一道二叉树的深度优先搜索的题目，在叶结点时将从根到叶结点的路径上的结点的值组成一个十进制的数，本质上还是一道深度优先搜索的题，只是换了一种考察的形式。这道题不难，只要细心点就可以bugfree的。

三.实现代码

#include <iostream>/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/struct TreeNode{    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution{    void sumNumbers(TreeNode*root, int &Result, int &TmpResult)    {        if (!root)        {            return;        }        // update the TmpResult        TmpResult = TmpResult * 10 + root->val;        if (!root->left && !root->right)        {            Result += TmpResult;        }        if (root->left)        {            sumNumbers(root->left, Result, TmpResult);        }        if (root->right)        {            sumNumbers(root->right, Result, TmpResult);        }        // restore the TmpResult;        TmpResult = (TmpResult - root->val) / 10;    }public:    int sumNumbers(TreeNode* root)    {        int Result = 0;        int TmpResult = 0;        sumNumbers(root, Result, TmpResult);        return Result;    }};

四.体会

    这道题是一道二叉树深度优先搜索的变形题目，考察的还是二叉树的递归遍历，只是换了一种考察方式而已，并没有多大新意。

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