112. 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.

Note: A leaf is a node with no children.

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.

思路

这是一道典型的DFS的题目，利用递归，不停的寻找左右子节点

退出条件：在某一个叶子节点（左右子节点为空），sum每一步减剩后的值等于叶子节点的val

C++11

/**
 * 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) {
        if (root == NULL) return false;
        if ((root->left == NULL) and (root->right == NULL) and (sum - root->val == 0)) return true;
        return hasPathSum(root->left, sum - root->val) or hasPathSum(root->right, sum - root->val);
    }
};