112. Path Sum

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#tree #leetcode

本文探讨了二叉树路径和问题的两种解决方案：递归方法和迭代方法（前序遍历）。通过实例讲解了如何确定从根节点到叶子节点的路径是否等于给定的总和，分析了递归方法的易错点，并详细比较了两种方法的时间复杂度和空间复杂度。

112. Path Sum

方法1: recursion
- 易错点
- Complexity
方法2: iterative（preorder）

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.

Note: A leaf is a node with no children.

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.

方法1: recursion

易错点

将generally 的终止base放在叶节点而不是null。一般情况下 null的时候可以判断，但如果是[], 0应该放回false。

Complexity

Time complexity : we visit each node exactly once, thus the time complexity isO(N), where NN is the number of nodes.
Space complexity : in the worst case, the tree is completely unbalanced, e.g. each node has only one child node, the recursion call would occur NN times (the height of the tree), therefore the storage to keep the call stack would be O(N). But in the best case (the tree is completely balanced), the height of the tree would be log(N). Therefore, the space complexity in this case would be O(log(N)).

class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if (!root) {
            return false;
        }
        if (!root -> left && !root -> right){
            return root -> val == sum;
        }
        
        return hasPathSum(root -> left, sum - root -> val) || hasPathSum(root -> right, sum - root -> val);
    }
};