[LeetCode]129.Sum Root to Leaf Numbers

【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.

【分析】

每到一个叶子节点就代表一条路径的结束，一个值的产生。累加每个值。

【代码】

/**********************************   日期：2014-12-30*   作者：SJF0115*   题目: 129.Sum Root to Leaf Numbers*   来源：https://oj.leetcode.com/problems/sum-root-to-leaf-numbers/*   结果：AC*   来源：LeetCode*   博客：*   时间复杂度：O(n)*   空间复杂度：O(logn)**********************************/#include <iostream>#include <queue>#include <vector>using namespace std;// 二叉树节点struct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public:    int sumNumbers(TreeNode *root) {        if(root == NULL){            return 0;        }//if        return SumNumbers(root,0);    }private:   int SumNumbers(TreeNode* node,int curSum){       if(node == NULL){           return 0;       }//if       curSum = curSum*10 + node->val;       // 到达叶子节点返回该路径上的值       if(node->left == NULL && node->right == NULL){           return curSum;       }       // 左子树       int leftSum = SumNumbers(node->left,curSum);       // 右子树       int rightSum = SumNumbers(node->right,curSum);       return leftSum + rightSum;   }};// 创建二叉树TreeNode* CreateTreeByLevel(vector<int> num){    int len = num.size();    if(len == 0){        return NULL;    }//if    queue<TreeNode*> queue;    int index = 0;    // 创建根节点    TreeNode *root = new TreeNode(num[index++]);    // 入队列    queue.push(root);    TreeNode *p = NULL;    while(!queue.empty() && index < len){        // 出队列        p = queue.front();        queue.pop();        // 左节点        if(index < len && num[index] != -1){            // 如果不空创建一个节点            TreeNode *leftNode = new TreeNode(num[index]);            p->left = leftNode;            queue.push(leftNode);        }        index++;        // 右节点        if(index < len && num[index] != -1){            // 如果不空创建一个节点            TreeNode *rightNode = new TreeNode(num[index]);            p->right = rightNode;            queue.push(rightNode);        }        index++;    }//while    return root;}int main() {    Solution solution;    // -1代表NULL    vector<int> num = {1,2,3,4,-1,-1,5};    TreeNode* root = CreateTreeByLevel(num);    cout<<solution.sumNumbers(root)<<endl;}

【思路二】

层次遍历

/*---------------------------------------*   日期：2015-05-08*   作者：SJF0115*   题目: 129.Sum Root to Leaf Numbers*   网址：https://leetcode.com/problems/sum-root-to-leaf-numbers/*   结果：AC*   来源：LeetCode*   博客：-----------------------------------------*/#include <iostream>#include <vector>#include <queue>using namespace std;// 二叉树节点struct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public:    int sumNumbers(TreeNode* root) {        if(root == nullptr){            return 0;        }//if        queue<TreeNode*> nodeQueue;        queue<int> sumQueue;        nodeQueue.push(root);        sumQueue.push(0);        int curSum = 0;        int totalSum = 0;        TreeNode* node;        // 层次遍历        while(!nodeQueue.empty()){            // 当前节点            node = nodeQueue.front();            nodeQueue.pop();            // 当前路径和            curSum = sumQueue.front();            sumQueue.pop();            curSum = curSum * 10 + node->val;            // 左子结点            if(node->left){                nodeQueue.push(node->left);                sumQueue.push(curSum);            }//if            // 右子结点            if(node->right){                nodeQueue.push(node->right);                sumQueue.push(curSum);            }//if            // 叶子节点计算总和            if(node->left == nullptr && node->right == nullptr){                totalSum += curSum;            }//if        }//while        return totalSum;    }};

层次遍历，对与每个节点都要记住父节点的当前和。因为计算每个节点的当前和都会因父节点而不一样。

到达叶子节点代表一条路径的结束。这个当前和加入到总和中。

【思路三】

先序遍历的非递归版本

/*---------------------------------------*   日期：2015-05-08*   作者：SJF0115*   题目: 129.Sum Root to Leaf Numbers*   网址：https://leetcode.com/problems/sum-root-to-leaf-numbers/*   结果：AC*   来源：LeetCode*   博客：-----------------------------------------*/#include <iostream>#include <vector>#include <queue>#include <stack>using namespace std;// 二叉树节点struct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public:    int sumNumbers(TreeNode* root) {        if(root == nullptr){            return 0;        }//if        stack<TreeNode*> nodeStack;        nodeStack.push(root);        stack<int> sumStack;        sumStack.push(0);        TreeNode* node;        int curSum = 0;        int totalSum = 0;        // 先序遍历 非递归        while(!nodeStack.empty()){            // 当前节点            node = nodeStack.top();            nodeStack.pop();            // 当前路径和            curSum = sumStack.top();            sumStack.pop();            curSum = curSum * 10 + node->val;            // 右子节点            if(node->right){                nodeStack.push(node->right);                sumStack.push(curSum);            }//if            // 左子结点            if(node->left){                nodeStack.push(node->left);                sumStack.push(curSum);            }//if            // 叶子节点计算总和            if(node->left == nullptr && node->right == nullptr){                totalSum += curSum;            }//if        }//while        return totalSum;    }};

【温故】

/*---------------------------------------*   日期：2015-05-08*   作者：SJF0115*   题目: 129.Sum Root to Leaf Numbers*   网址：https://leetcode.com/problems/sum-root-to-leaf-numbers/*   结果：AC*   来源：LeetCode*   博客：-----------------------------------------*/#include <iostream>#include <vector>#include <queue>using namespace std;// 二叉树节点struct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public:    int sumNumbers(TreeNode* root) {        if(root == nullptr){            return 0;        }//if        int curSum = 0;        int totalSum = 0;        helper(root,curSum,totalSum);        return totalSum;    }private:    void helper(TreeNode* root,int curSum,int &totalSum){        if(root == nullptr){            return;        }//if        // 先序遍历        curSum = curSum * 10 + root->val;        // 在叶子节点处计算总和        if(root->left == nullptr && root->right == nullptr){            totalSum += curSum;            return;        }//if        // 左子树        helper(root->left,curSum,totalSum);        // 右子树        helper(root->right,curSum,totalSum);    }};