112. Path Sum (Tree; DFS)

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

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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(!root) return false;
        flag = false;
        target = sum;
        preOrder(root,0);
        return flag;
        
    }
    void preOrder(TreeNode* node,int sum){
        sum = node->val + sum; //递归前，加上当前节点
        if(node->left)
        {           
            preOrder(node->left,sum);
        }
        if(node->right)
        {           
            preOrder(node->right,sum);
        }
        if(!node->left && !node->right && sum == target) //递归结束条件：到了叶子节点
        {
            flag = true;
        }
    }
private:
    bool flag;
    int target;
};

 

转载于:https://www.cnblogs.com/qionglouyuyu/p/4854200.html