LeetCode: Path Sum [112]

【题目】

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.

【题意】

判断二叉树中是否存在一条从根到叶子节点的路径，使得路径上的节点值之和等于所给的值

【思路】

    DFS，递归

【代码】

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    
    bool dfs(TreeNode*root, int pathSum, int sum){
        pathSum+=root->val;
        if(root->left==NULL && root->right==NULL){
            return pathSum==sum;
        }
        
        //搜索左子树
        bool hasLeft=false;
        if(root->left)
            hasLeft=dfs(root->left, pathSum, sum);
        //搜索右子树
        bool hasRight=false;
        if(root->right)
            hasRight=dfs(root->right, pathSum, sum);
            
        return hasLeft || hasRight;
        
    }

    bool hasPathSum(TreeNode *root, int sum) {
        if(root==NULL)return false;
        int pathSum=0;
        return dfs(root, pathSum, sum);
    }
};