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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. 
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.

分析

题意是给定一颗二叉树和一个和sum，判断这颗二叉树是否有一条root-to-leaf的路径，使经过的节点的val和等于sum.

、 
     * 树的很多问题都可以用递归来解决，这道题目找的是root-to-leaf的和，那么root是肯定经过的
     * 所以我们可以分解出树的子树来解决:
     * 1、如果当前节点是叶子节点，而且当前节点的val等于sum，那么返回true
     * 2、对于其他情况，递归调用自身的左子树和右子树来解决；

Code

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    /* 
     * 树的很多问题都可以用递归来解决，这道题目找的是root-to-leaf的和，那么root是肯定经过的
     * 所以我们可以分解出树的子树来解决:
     * 1、如果当前节点是叶子节点，而且当前节点的val等于sum，那么返回true
     * 2、对于其他情况，递归调用自身的左子树和右子树来解决；
     * */
    bool hasPathSum(TreeNode *root, int sum) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if( !root )     return false;
        if(root->val == sum && root->left == NULL && root->right == NULL)
            return true;
        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->right);
    }
};

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