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Path Sum⚓︎

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Description⚓︎

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:

  • 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.

example 1

Example 2:

  • Input: root = [1,2,3], targetSum = 5
  • Output: false
  • Explanation:
    • There two root-to-leaf paths in the tree:
      • (1 --> 2): The sum is 3.
      • (1 --> 3): The sum is 4.
    • There is no root-to-leaf path with sum = 5.

Example 3:

  • Input: root = [], targetSum = 0
  • Output: false
  • Explanation: Since the tree is empty, there are no root-to-leaf paths.

Constraints:

  • The number of nodes in the tree is in the range [0, 5000].
  • -1000 <= Node.val <= 1000
  • -1000 <= targetSum <= 1000

Solution⚓︎

Way 1 (Recursion)⚓︎

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/**
 * 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 hasPathSum(TreeNode* root, int targetSum) {
        if (!root) return false;
        if (!root->left && !root->right && targetSum == root->val) return true;
        return hasPathSum(root->left, targetSum - root->val) || hasPathSum(root->right, targetSum - root->val);
    }
};
  • Time complexity: \(O(N)\), where \(N\) is the number of nodes in the tree;
  • Space complexity: \(O(H)\), where \(H\) is the height of the tree (Worst: \(O(N)\); Best: \(O(\log N)\) ).

Way 2 (Pre-order)⚓︎

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class Solution {
public:
    bool hasPathSum(TreeNode* root, int targetSum) {
        if (!root) return false;
        stack<pair<TreeNode*, int>> stk;
        stk.push(make_pair(root, root->val));
        while (!stk.empty()) {
            auto node = stk.top();
            stk.pop();
            if (!(node.first)->left && !(node.first)->right && targetSum == node.second) return true;
            if ((node.first)->right)
                stk.push(make_pair((node.first)->right, node.second + (node.first)->right->val));
            if ((node.first)->left)
                stk.push(make_pair((node.first)->left, node.second + (node.first)->left->val));
        }
        return false;
    }
};

Way 3 (Level-order)⚓︎

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class Solution {
public:
    bool hasPathSum(TreeNode* root, int targetSum) {
        if (!root) return false;
        queue<pair<TreeNode*, int>> q;
        q.push(make_pair(root, root->val));
        while (!q.empty()) {
            auto t = q.front();
            q.pop();
            if (!t.first->left && !t.first->right && t.second == targetSum) return true;
            if (t.first->left) q.push(make_pair(t.first->left, t.first->left->val + t.second));
            if (t.first->right) q.push(make_pair(t.first->right, t.first->right->val + t.second));
        }
        return false;
    }
};
  • Time complexity: \(O(N)\);
  • Space complexity: \(O(N)\).