Sum Root to Leaf Numbers

Sum Root to Leaf Numbers

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Question Solution 

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.

这道题要我们找出从根节点到所有叶子节点的十进制数字的和，属于很基础的树的遍历题。

解法一：

使用DFS从根节点开始遍历树，在从上到下搜索过程中记录当前十进制数字，当到达一个叶子节点时，累加到总和中。

递归问题要注意何时结束递归。

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int sumNumbers(TreeNode *root) {        sum = 0;        recursiveSum(root, 0);        return sum;    }private:    int sum;    void recursiveSum(TreeNode *root, int curSum) {        if (NULL == root) {            return;        }        curSum = curSum * 10 + root->val;        if (NULL == root->left && NULL == root->right) {            sum += curSum;        }        recursiveSum(root->left, curSum);        recursiveSum(root->right, curSum);    }};

解法二：

这样的遍历问题也可以使用非递归的方法，常用C++ queue容器存储队列元素，先进先出，从根节点遍历过程中按层次把

节点信息加入队列中，要注意到达叶子节点的操作。

/** * Definition for binary tree * 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 (NULL == root) {            return 0;        }        int sum = 0;        queue<NodeInfo> q;        q.push(NodeInfo(root->val, root));                while (!q.empty()) {            NodeInfo ni = q.front();            q.pop();            // when it is a left node            if (NULL == ni.nodePtr->left && NULL == ni.nodePtr->right) {                sum += ni.pathNum;                continue;            }            if (ni.nodePtr->left != NULL) {                q.push(NodeInfo(ni.pathNum * 10 + ni.nodePtr->left->val, ni.nodePtr->left));            }            if (ni.nodePtr->right != NULL) {                q.push(NodeInfo(ni.pathNum * 10 + ni.nodePtr->right->val, ni.nodePtr->right));            }        }        return sum;    }private:    typedef struct NodeInfo {        int pathNum;        TreeNode *nodePtr;        NodeInfo(int _pathNum, TreeNode *_nodePtr) {            pathNum = _pathNum;            nodePtr = _nodePtr;        }    }NodeInfo;};