112. Path Sum

Leetcode Tree Depth-first Search

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.

分析

如果是叶子节点，那么路径的和要刚好等于给定的和。如果不是叶子节点，那么递归求解该节点的左右子节点，子节点和相应的减少该节点的值。

public boolean hasPathSum(TreeNode root, int sum) {
    if (root == null) return false;
    if (root.left == null && root.right == null)  return sum == root.val; //叶子节点
    int newSum = sum - root.val;
    return hasPathSum(root.left, newSum) || hasPathSum(root.right, newSum);  
}