Sum Root to Leaf Numbers
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/** * 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) { int sum=0; sumpath(root,0,sum); return sum; } void sumpath(TreeNode *p, int presum, int& sum ) { if(NULL==p) return; presum=presum*10+p->val; if(p->left==NULL && p->right==NULL) { sum+=presum; } sumpath(p->left,presum,sum); sumpath(p->right,presum,sum); }};树的递归两种策略:
1. 结果在root产生,可以遵循traversal order, 一般函数递归返回结果
2. 结果在leaf节点产生,并难以遵循traversal order, 并且产生结果需要父节点的数据,可以设置全局变量或者变量引用,在递归中变量得到更新,最终返回这个变量(如这题) 0 0
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
- Sum Root to Leaf Numbers
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