Q:
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.
Analysis:
题目要求给出路径和,判断二叉树根到叶子结点是否有累加和等于路径和的分支。
采用递归方法,依次递归每个分支,直到有相等的分值为止。
Code:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null) {
return false;
} else if (root.left == null && root.right == null && root.val == sum) {// 递归结束条件
return true;
} else {
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}
}
}