leetcode 112. Path Sum(C語言,二叉樹,遞迴思想)28
阿新 • • 發佈:2019-01-26
貼原題:
Given a binary tree and a sum, determine if the tree has aroot-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,
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
解析
本題是給出一個二叉樹和一個目標值,讓判斷是否存在一個叉樹使得從根節點到葉節點所有的數之和等於給定的目標值,存在則返回true,否則返回false。
那麼我的思路是利用遞迴思想,逐級判斷,直到一個葉節點,如果其累加和等於目標值則返回true,否則返回false。葉節點的特徵是父節點存在,左右孩子節點不存在。
貼程式碼:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
bool ifEqualTarget(struct TreeNode* root, int target, int sum)
{
if(root!=NULL && root->left==NULL && root->right==NULL && root->val+sum==target)//葉節點
{
return true;
}
if(root!=NULL)
{
if(ifEqualTarget(root->left, target, root-> val+sum))//左孩子遞迴
{
return true;
}
if(ifEqualTarget(root->right, target, root->val+sum))//右孩子遞迴
{
return true;
}
}
return false;
}
bool hasPathSum(struct TreeNode* root, int sum) {
return root!=NULL && ifEqualTarget(root, sum, 0);//首先得滿足根節點存在
}