LeetCode 112. 路径总和 递归 树的遍历

给定一个二叉树和一个目标和，判断该树中是否存在根节点到叶子节点的路径，
这条路径上所有节点值相加等于目标和。

说明: 叶子节点是指没有子节点的节点。

示例: 
给定如下二叉树，以及目标和 sum = 22，

              5
             /             4   8
           /   /           11  13  4
         /  \              7    2      1
返回 true, 因为存在目标和为 22 的根节点到叶子节点的路径 5->4->11->2。

/**
 * 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 dfs(TreeNode* root, int curr, int sum)
{
    bool ret1 = false; bool ret2 = false;
    curr += root->val;
    if(root->left == NULL && root->right == NULL)   return  curr == sum;
    if(root->left!= NULL) ret1 = dfs(root->left,curr,sum);
    if(root->right != NULL) ret2 = dfs(root->right,curr,sum);

    return ret1 || ret2;
}

bool hasPathSum(TreeNode* root, int sum) {
    if (root ==  NULL) return false;
    int curr = 0;bool ret1 = false,ret2 = false;
    return dfs(root,curr,sum);
}

};

class Solution {
public:

bool dfs(TreeNode* root, int curr, int sum)
{
    if(root == NULL) return false;
    bool ret1 = false; bool ret2 = false;
    curr += root->val;
    if(root->left == NULL && root->right == NULL)   return  curr == sum;
    return dfs(root->left,curr,sum) || dfs(root->right,curr,sum);
}

bool hasPathSum(TreeNode* root, int sum) {
    int curr = 0;
    return dfs(root,curr,sum);
}

};