leetcode——129——Sum Root to Leaf Numbers

最新推荐文章于 2019-03-17 23:22:15 发布

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本文介绍了如何使用递归和非递归方法解决求二叉树从根节点到叶子节点的路径所表示的数字之和的问题。详细解释了两种方法的实现过程，并通过实例演示了求和的具体步骤。

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.

递归解法：

/**
 * 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:
	int sumNumbers(TreeNode* root) {
        return dfs(root, 0);
	}
	int dfs(TreeNode* root,int sum)
	{
	     
         if(root == NULL) return 0;
         if(root->left == NULL && root->right == NULL) return root->val + sum * 10;         
         return dfs(root->left, root->val + sum * 10) + dfs(root->right, root->val + sum * 10); 

	}
	
};

非递归解法：

/**
 * 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:
	int sumNumbers(TreeNode* root) {
	    int sum = 0;
        if(root == NULL) 
            return sum;
        queue<TreeNode*> queue; 
        queue.push(root);
        while(!queue.empty()) {
            int levelLen = queue.size();            
            for (int i = 0; i < levelLen; i++) {
                TreeNode* node = queue.front();
                queue.pop();
                if(node->left == NULL && node->right == NULL)
                    sum += node->val;
                if(node->left != NULL) {
                    node->left->val += node->val * 10;
                    queue.push(node->left);
                }
                if(node->right != NULL) {
                    node->right->val += node->val * 10;
                    queue.push(node->right);
                }
            }
        }
        return sum;
    }
	
};