112. Path Sum

Link

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.

Solution

假設選取目前節點，縮減sum成 sum-node.val條件的情況下看看左右子樹是否有其中一個可以符合。 1. 要記得edge case的判斷 2. 舉簡單的例子看看是否work

T: O(N), 基本上也是每個node都要判斷，左子樹or右子樹的case

class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
        if(root.left == null && root.right == null && root.val == sum){
            return true;
        }
        else{
            return hasPathSum(root.left, sum-root.val) || hasPathSum(root.right, sum-root.val);
        }
        
    }
}