leetcode sum path II

Binary tree? And the path!

Title
Given a binary tree and a target and to find all paths from the root node to the leaf node is equal to the sum of the path and the given target.
Description: leaf node is a node has no child nodes.

Examples
Example:
Given the following binary tree, and the target and sum = 22,

          5
         / \
        4   8
       /   / \
      11  13  4
     /  \    / \
    7    2  5   1

Returns:
[
[5,4,11,2],
[5,8,4,5]
]

Topic analysis

  • Goal: to find all paths starting from the root to leaf node ⇒ its junction equals the target value
  1. Find the path from the root to leaf node ⇒ recursive solution
  2. If the node is equal to the sum of the target value path ⇒ Add

Problem-solving ideas

variable effect
find() Seeking the path function

process

  1. ⇒ Add current node to update the path of the node and
  2. Analyzing the node is equal to the sum of the target value and the current node is a leaf node if the condition is satisfied ⇒ add the path to the answer
  3. If the left subtree is not empty ⇒ update left child node to the current node, repeat 2,3,4
  4. If the right subtree is not empty ⇒ update right child nodes for the current node, repeat 2,3,4

code show as below

void find(TreeNode*root, int add, int sum, vector<int> path, vector<vector<int>> &ans)
 {
     add += root->val;                             //更新结点和
     path.push_back(root->val);                    //当前结点加入路径
     if(add == sum&&!root->left&&!root->right) ans.push_back(path);  
        if(root->left) find(root->left,add,sum,path,ans);
        if(root->right) find(root->right,add,sum,path,ans);
 } 
class Solution {
public:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        if(!root) return{};
        vector<vector<int>> ans;
        vector<int> path;
        int add =0;
        find(root, add, sum, path, ans);
        return ans;
    }
};
Published 34 original articles · won praise 0 · Views 576

Guess you like

Origin blog.csdn.net/Luyoom/article/details/104860955