leetcode:path sum(I) 递归与非递归解法

最新推荐文章于 2020-07-08 16:01:18 发布

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本文探讨了二叉树中寻找从根节点到叶子节点的路径，使得路径上所有节点值之和等于给定值的问题。提供了两种解决方案：递归方法和非递归方法，并详细解释了每种方法的具体实现。

Path Sum

Total Accepted: 12663 Total Submissions: 42061 My Submissions

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.

递归解法：

class Solution{

public:

bool hasPathSum(TreeNode *root, int sum) {

if(!root)

return false;

else if(root->val == sum && !root->left && !root->right)

return true;

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

};

非递归解法：定义两个栈，一个存放前序遍历的结点，一个存放前序遍历时的结点累加值；

/**
* Definition for binary tree
* 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)
return false;
stack<TreeNode *> stk;
stack<int> tempSum;

stk.push(root);
tempSum.push(root->val);
while(!stk.empty())
{
TreeNode *nd = stk.top();
stk.pop();
int tpm = tempSum.top();
tempSum.pop();
if(nd->left == NULL && nd->right == NULL && tpm == sum)
{
return true;
}
if(nd->left)
{
stk.push(nd->left);
tempSum.push(tpm + nd->left->val);
}
if(nd->right)
{
stk.push(nd->right);
tempSum.push(tpm + nd->right->val);
}
}
return false;
}
};