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0112. Path Sum

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

Example 2:

img

Input: root = [1,2,3], targetSum = 5
Output: false

Example 3:

Input: root = [1,2], targetSum = 0
Output: false

Constraints:

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

Analysis

The path has be from root to leaf, and we need to make sure our code satisfy both:

  1. if node->left==nullptr && node->right==nullptr, then the node is a leaf node
  2. if current sum==sum, then current path sum is satisfied.

We can then use recursion to solve.

  • TIme: O(n) where n is number of nodes in the tree
  • Space: O(n) worst case when tree is a linkedlist

Code

/**
 * 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 sum) {
        if (!root) return false;// already visited to the children node of leaf
        sum -= root -> val;
        // is leaf node + curr sum == target sum
        if (!sum && !root -> left && !root -> right) return true;
        return hasPathSum(root -> left, sum) || hasPathSum(root -> right, sum);
    }
};
Last update: April 1, 2022