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No. 112 - Path Sum

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LeetCode 112 Path Sum

題目描述

Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

A leaf is a node with no children.

Example 1.

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Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output: true
Explanation: The root-to-leaf path with the target sum is shown.

解法

本題的目的是要尋找一顆 tree 從 root 到任一 leaf 的所有 nodes 的和是否有與 targetSum 相等。

從這題目的敘述我們可以想到的是就是用 depth first search 來解,用 DFS 來計算每一條 root-to-leaf 路徑的 nodes 總和,並在每條 path 走到最後的 node 加總完後,跟 targetSum 進行比較確認是否相同。

完整的程式如下:

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: bool dfs(TreeNode *node,int pathSum, int targetSum) { if (!node) return false; if (!node->left && !node->right) return pathSum + node->val == targetSum; pathSum += node->val; return dfs(node->left, pathSum, targetSum) || dfs(node->right, pathSum, targetSum); } bool hasPathSum(TreeNode* root, int targetSum) { return dfs(root, 0, targetSum); } };

我們宣告一個 dfs 函式來進行 DFS,每一次的 recursive 我們需要知道當前走訪到的 node 以及前面走過的 nodes 的總和 pathSum,同時也需要把我們最後需要比較的 targetSum 一路往下傳。

程式第 17 行當判別出 DFS 已經走到 leaf 則會跟 targetSum 進行比較確認是否相等。

程式第 20 行則是 dfs recursive 的做法,繼續走訪當前 node 的 left 跟 right child 直到走到 leaf。而最後我們只需要用 logic OR 來判斷是否存在一條 path 的總和與 targetSum 相等即可。

複雜度分析

我們需要走訪每個 nodes 所以時間複雜度為

O(N)

我們需要存 dfs 的函式 stack 空間複雜度為

O(H)