129. Sum Root to Leaf Numbers

最新推荐文章于 2020-12-22 20:22:53 发布

ShenYounger

最新推荐文章于 2020-12-22 20:22:53 发布

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分类专栏： leetcode

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本文介绍了一种解决二叉树中所有从根节点到叶子节点路径数字之和的方法。通过非递归的方式，利用队列存储节点及其路径值，最终求得总和。

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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.

分析：简单的树的遍历，下面代码使用队列，没有使用递归，队列节点除了存储节点地址还存储了从根节点到该节点路径表示的值。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    struct Qnode
    {
      int sum_;
      TreeNode* p_;
      Qnode(int s, TreeNode *p):sum_(s),p_(p){}
      Qnode(){}
    };
    int sumNumbers(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(root == NULL)return 0;
        queue<Qnode> myqueue;
        myqueue.push(Qnode(root->val, root));
        int res = 0;
        while(myqueue.empty() == false)
        {
            Qnode tmp = myqueue.front();
            myqueue.pop();
            if(tmp.p_->left == NULL && tmp.p_->right == NULL)
                res += tmp.sum_;
            else
            {
                if(tmp.p_->left != NULL)
                    myqueue.push(Qnode(tmp.sum_*10 + tmp.p_->left->val, tmp.p_->left));
                if(tmp.p_->right != NULL)
                    myqueue.push(Qnode(tmp.sum_*10 + tmp.p_->right->val, tmp.p_->right));
            }
        }
        return res;
    }
};