二叉树的路径总和

给你二叉树的根节点 root 和一个表示目标和的整数 targetSum 。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 targetSum 。如果存在，返回 true ；否则，返回 false 。

采用递归的方法

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */


bool hasPathSum(struct TreeNode* root, int targetSum) {
    if(root==NULL)
    return false;
    if(root->left==NULL&&root->right==NULL)
    return targetSum==root->val;
    return hasPathSum(root->left,targetSum-root->val)||hasPathSum(root->right,targetSum-root->val);
    
}

提供python的写法

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
        if not root:
            return False
        if not root.left and not root.right:
            return targetSum==root.val
        return self.hasPathSum(root.left,targetSum-root.val) or self.hasPathSum(root.right,targetSum-root.val)