leetcoder-112-Path Sum

Path Sum

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.

是否存在从根节点到叶子节点的和，等于所给的数。

递归 one

/**
 * 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 solve(TreeNode* root,int sum,bool& f,int s){
        if(!root->left&&!root->right){
            if(s==sum) f=true;
            return;
        }
        if(root->left) solve(root->left,sum,f,s+root->left->val);
        if(root->right) solve(root->right,sum,f,s+root->right->val);
    }
    bool hasPathSum(TreeNode* root, int sum) {
        bool f=false;
        if(!root) return false;// 为空  则返回false

        solve(root,sum,f,root->val);
        return f;
    }
};

递归 two

/**
 * 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:
    bool hasPathSum(TreeNode* root, int sum) {
        return solve(root,sum,0);
    }
    bool solve(TreeNode* root,int sum,int s){
        if(!root) return false;
        if(!root->left&&!root->right)
             return sum==s+root->val;

        return solve(root->left,sum,s+root->val)||solve(root->right,sum,s+root->val);//存在即可 故用 ||
    }
};

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