Sum Root to Leaf Numbers

Sum Root to Leaf Numbers

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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.

這道題要我們找出從根節點到所有葉子節點的十進制數字的和，屬於很基礎的樹的遍歷題。

解法一：

使用DFS從根節點開始遍歷樹，在從上到下搜索過程中記錄當前十進制數字，當到達一個葉子節點時，累加到總和中。

遞歸問題要注意何時結束遞歸。

/**
 * 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：

這樣的遍歷問題也可以使用非遞歸的方法，常用C++ queue容器存儲隊列元素，先進先出，從根節點遍歷過程中按層次把

節點信息加入隊列中，要注意到達葉子節點的操作。

/**
 * 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;
};