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); }
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