112. Path Sum （判断路径和是否等于某值)

 

 

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

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

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

 

 

 

 

class Solution {
public:
    
    bool hasPathSum(TreeNode* root, int targetSum) {
        if(root == nullptr)  return false;

        if (root->left == nullptr && root->right == nullptr && targetSum == root->val) return true;
        
        return hasPathSum(root->left,targetSum-root->val) || hasPathSum(root->right,targetSum-root->val);
    }
};

 

 

class Solution {
public:
    bool res = false;
    void backtrack(TreeNode* root, int cursum, int targetSum) {
        if (root == nullptr) return;
        if(root->left == nullptr && root->right == nullptr) {
            // 需要加上叶子节点
            if (cursum + root->val == targetSum) {
                res = true;
                return;
            }
        }
        // 剪枝
        if(!res) backtrack(root->left,cursum+root->val,targetSum);
        if(!res) backtrack(root->right,cursum+root->val,targetSum);
    }
    
    bool hasPathSum(TreeNode* root, int targetSum) {
        if (root == nullptr) return res;
        backtrack(root,0,targetSum);
        return res;
    }
};