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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.
思路：最近睡得不好，搞得脑子也短路了。一直想着怎么从叶节点往上获取数字。费了好大的劲才AC了，结果一看答案，立马郁闷了。不到10行就能搞定。
具体的原理是深度优先，在向下迭代的过程中，用一个变量不断的把父节点以上的数字传下来，返回的时候把结果加起来。
public int sumNumbers(TreeNode root) { return helper(root, 0); } public int helper(TreeNode root, int curSum) { if(root == null) return 0; curSum = curSum*10 + root.val; if(root.left == null && root.right == null) return curSum; return helper(root.left, curSum) + helper(root.right, curSum); }
下面是我自己AC的代码，超长，超繁琐，当个反面例子吧。思路是把每个数字都存在一个vector里面。之所以繁琐是因为我只是在返回的时候下功夫。
int sumNumbers(TreeNode *root) { int s, e, sum = 0; vector<vector<int>> v; Numbers(root, s, e, v); for(vector<vector<int>>::iterator it = v.begin(); it != v.end(); ++it) { int num = 0; while(!it->empty()) { num = num * 10 + it->back(); it->pop_back(); } sum += num; } return sum; } void Numbers(TreeNode * root, int &s, int &e, vector<vector<int>>& v) { int sl = -1, el = -1, sr = -1, er = -1; s = e = -1; if(root == NULL) return; if(root->left == NULL && root->right == NULL) //在叶节点 压入新的数字 { e = s = v.size(); v.push_back(vector<int>(1,root->val)); return; } if(root->left != NULL) { Numbers(root->left, sl, el, v); for(int i = sl; i <= el; ++i) { v[i].push_back(root->val); } s = sl; e = el; } if(root->right != NULL) { Numbers(root->right, sr, er, v); for(int i = sr; i <= er; ++i) { v[i].push_back(root->val); } s = (s == -1) ? sr : s; e = er; } }
【leetcode】Sum Root to Leaf Numbers（hard）
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原文地址：http://www.cnblogs.com/dplearning/p/4469468.html
