Sword refers to the path of the binary tree and a specific value in offer, and sword refers to the binary tree in offer.

Sword refers to the path of the binary tree and a specific value in offer, and sword refers to the binary tree in offer.

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Question link: http://www.nowcoder.com/practice/b736e784e3e34731af99065031301bca? Rp = 2 & ru =/ta/coding-interviews & qru =/ta/coding-interviews/question-ranking

Description
Enter a binary tree and an integer to print the sum of the node values in the binary tree and all the paths of the input integers. The path is defined as a path formed from the root node of the tree down to the node through which the leaf node passes.

Ideas

We start a level-1 search from the root node and press the node into the path for each search. Then, we can determine whether the node is a leaf node. If it reaches the leaf node, it determines whether the sum of the path meets the conditions. If the conditions are met, the path is saved.

Once we reach the bottom layer, we still need to trace back layer by layer, so we need to use the features of Deep Search and backtracking.

/*struct TreeNode {int val;struct TreeNode *left;struct TreeNode *right;TreeNode(int x) :val(x), left(NULL), right(NULL) {}};*/class Solution{public:vector<vector<int> > FindPath(TreeNode* root,int expectNumber){if(root==nullptr) return ans;int sum = 0;vector<int> path;Find(root,path,sum,expectNumber);return ans;}void Find(TreeNode *root,vector<int> path,int sum,int expectNumber){sum+=root->val;path.push_back(root->val);bool isLeaf = (root->left==nullptr && root->right==nullptr);if(sum==expectNumber && isLeaf){ans.push_back(path);}if(root->left!=nullptr)Find(root->left,path,sum,expectNumber);if(root->right!=nullptr)Find(root->right,path,sum,expectNumber);path.pop_back();}private:vector<vector<int> > ans;};

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