Leetcode[129]-Sum Root to Leaf Numbers

Link: https://leetcode.com/problems/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.

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这道题让我真的醉了，先说下思路吧。

先求出每个叶子节点的值，然后将这些值相加。求叶子节点的值可以通过深度优先遍历，一遍往下，一遍增加各个节点的值，如果是叶子节点就将数组值的和加入到一个只有叶子节点的数组中，最后求这个只有叶子节点的数组的值得和。从头到尾自己写的代码，可以AC了，还没有进行优化，先做到这里吧。

代码（C++）：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    //return sum of newleafValue    int sumNumbers(TreeNode* root) {        if(root == NULL) return 0;        if(!root->left && !root->right) return root->val;        vector<int> nums = saveLeafNums(root);        return sumV(nums);    }    //save leftnode to vector    vector<int> saveLeafNums(TreeNode * root){        vector<int> nums,nodeValue;        int eachSum = 0;        stack<TreeNode *> stk;        map<TreeNode *, int> visited;        TreeNode *p = root;        stk.push(p);        visited[p] = 0;        nodeValue.push_back(p->val);        while(!stk.empty()){            while(p->left && visited[p]==0 ){                //if this node is visited,break                 if(visited[p->left]==1) break;                p = p->left;                stk.push(p);                visited[p] = 0;                tenMutilVector(nodeValue);//enlarge vector element's value                nodeValue.push_back(p->val);            }            if(!stk.empty()) {                p = stk.top();                if(!p->left && !p->right){                    nums.push_back(sumV(nodeValue));                }                if(p->right && (visited.find(p->right)==visited.end() || visited[p->right]==0)){                    p = p->right;                    stk.push(p);                    visited[p] = 0;                    tenMutilVector(nodeValue);                    nodeValue.push_back(p->val);                    continue;                }                visited[p] = 1;                stk.pop();                tenDivideVector(nodeValue);                nodeValue.resize(nodeValue.size()-1);            }        }        return nums;    }    void tenMutilVector(vector<int> &nums) {        int n = nums.size();        for(int i = 0; i < n; i++){            nums[i] *= 10;        }    }    void tenDivideVector(vector<int> &nums) {        int n = nums.size();        for(int i = 0; i < n; i++){            nums[i] /= 10;        }    }    int sumV(vector<int> &nums){        int n = nums.size();        int sumValue = 0;        for(int i = 0; i < n; i++){            sumValue += nums[i];        }        return sumValue;    }   };

附加一个求深度的代码：

    int getDepth(TreeNode* root) {        if(root == NULL) return 0;        int ldepth = 0,rdepth = 0;        ldepth = getDepth(root->left);        rdepth = getDepth(root->right);        return ldepth>rdepth?ldepth+1:rdepth+1;    }

方法二：（递归法）

class Solution {public:    int sumNumbers(TreeNode* root) {        sum = 0;        sumStep(root, 0);        return sum;    }private:    int sum;    void sumStep(TreeNode *root, int num){        if(!root) return;        if(!root->left && !root->right) {            sum = sum + num*10 + root->val;        }else{            sumStep(root->left, num*10 + root->val);            sumStep(root->right, num*10 + root->val);        }    }};