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[Leetcode]-Path Sum

时间:2017-06-22 20:34:40      阅读:186      评论:0      收藏:0      [点我收藏+]

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.

Hide Tags :Tree, Depth-first Search
题目:找出二叉树中是否有从根节点到树叶路径的值之和等于给定的数据sum
思路:凡是设计到二叉树的查找,搜索…採用递归是一种较好的方法。
1、给出递归终止条件,也就是
   A:当节点为NULL时返回false
   B:当节点为树叶(无儿子节点)时,检查此树叶的val是否与sum相等,相等则返回true。否则返回false
2、当不是终止条件B时。继续向下递归。

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

bool hasPathSum(struct TreeNode* root, int sum) {

    if(root == NULL)    return false;
    if(root->left == NULL && root->right == NULL) return sum == root->val;

    return (hasPathSum(root->left,sum - root->val) || hasPathSum(root->right,sum - root->val));

}

[Leetcode]-Path Sum

原文:http://www.cnblogs.com/mfmdaoyou/p/7067007.html

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