112. Path Sum

2017-06-19  本文已影响0人  YellowLayne

1.描述

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.

2.分析

3.代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
bool hasPathSum(struct TreeNode* root, int sum) {
    if (NULL == root) return false;
    if (NULL == root->left && NULL == root->right) return root->val == sum;
    if (NULL == root->left) return hasPathSum(root->right, sum - root->val);
    if (NULL == root->right) return hasPathSum(root->left ,sum - root->val);
    bool left = hasPathSum(root->left, sum - root->val);
    bool right = hasPathSum(root->right, sum - root->val);
    return left || right;
}
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