LeetCode 437：路径总和 III 解法精讲

问题描述

LeetCode 437. 路径总和 III 要求统计二叉树中路径和等于目标值的路径数量。路径不需要从根节点开始，也不需要在叶子节点结束，但必须是从父节点指向子节点的方向。

解题思路

暴力递归法

遍历二叉树的每个节点，以该节点为起点向下搜索所有可能的路径，统计路径和等于目标值的数量。这种方法简单直观，但时间复杂度较高。

def pathSum(root, targetSum):
    if not root:
        return 0
    
    def dfs(node, current_sum):
        if not node:
            return 0
        current_sum += node.val
        count = 1 if current_sum == targetSum else 0
        count += dfs(node.left, current_sum)
        count += dfs(node.right, current_sum)
        return count
    
    return dfs(root, 0) + pathSum(root.left, targetSum) + pathSum(root.right, targetSum)

前缀和优化法

利用前缀和的思想优化时间复杂度。在遍历二叉树时，记录从根节点到当前节点的路径和，并通过哈希表快速查找是否存在满足条件的前缀和。

def pathSum(root, targetSum):
    from collections import defaultdict
    prefix_sum = defaultdict(int)
    prefix_sum[0] = 1
    
    def dfs(node, current_sum):
        if not node:
            return 0
        current_sum += node.val
        count = prefix_sum.get(current_sum - targetSum, 0)
        prefix_sum[current_sum] += 1
        count += dfs(node.left, current_sum)
        count += dfs(node.right, current_sum)
        prefix_sum[current_sum] -= 1
        return count
    
    return dfs(root, 0)

复杂度分析

暴力递归法

时间复杂度：O(n^2)，其中 n 是二叉树的节点数。最坏情况下，二叉树退化为链表。
空间复杂度：O(n)，递归栈的深度取决于树的高度。

前缀和优化法

时间复杂度：O(n)，每个节点只需遍历一次。
空间复杂度：O(n)，哈希表存储前缀和的数量。

边界条件与注意事项

空树直接返回 0。
目标值可能为负数，不能提前终止搜索。
哈希表需要初始化 prefix_sum[0] = 1，表示路径和为 0 的情况。

测试用例

# 示例 1
root = TreeNode(10)
root.left = TreeNode(5)
root.right = TreeNode(-3)
root.left.left = TreeNode(3)
root.left.right = TreeNode(2)
root.right.right = TreeNode(11)
root.left.left.left = TreeNode(3)
root.left.left.right = TreeNode(-2)
root.left.right.right = TreeNode(1)
targetSum = 8
assert pathSum(root, targetSum) == 3

总结

路径总和 III 是一个典型的二叉树路径问题，暴力递归法易于理解但效率较低，前缀和优化法通过空间换时间显著提升性能。掌握这两种方法有助于解决类似问题。

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