【leetcode】129. Sum Root to Leaf Numbers
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最重要的关键就是当只有单亲孩子的时候,不能重复计数。
所以需要对当前结点是否有双亲孩子,单亲孩子,还有叶节点的分别考虑。主要问题在于重复计数。
/** * @author johnsondu * @time 20:51 8th Oct 2015 * @type binary tree traverse * @url https://leetcode.com/problems/sum-root-to-leaf-numbers/ * @status Accepted *//** * 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: void dfs(int cnt, TreeNode* node, int &ans) { cnt = cnt * 10 + node->val; if(node->left && node->right) { dfs(cnt, node->left, ans); dfs(cnt, node->right, ans); } else if(node->left) dfs(cnt, node->left, ans); else if(node->right) dfs(cnt, node->right, ans); else ans += cnt; } int sumNumbers(TreeNode* root) { int ans = 0; if(root == NULL) return ans; int cnt = root->val; if(root->left && root->right) { dfs(cnt, root->left, ans); dfs(cnt, root->right, ans); } else if(root->left) dfs(cnt, root->left, ans); else if(root->right) dfs(cnt, root->right, ans); else ans += cnt; return ans; }};
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