112. Path Sum 二叉树路径的和

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.

For 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.

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public bool HasPathSum(TreeNode root, int sum) {
        return dfs(root, sum);
    }
    
	private bool dfs(TreeNode root, int sum) {
		if (root == null)
			return false;
		if (root.left == null && root.right == null && sum == root.val)
			return true;
		return dfs(root.left, sum - root.val) || dfs(root.right, sum - root.val);
	}
}

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