Path Sum

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 andsum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path5->4->11->2which sum is 22.

Analysis

递归求解,注意的是对path的定义:root-to-leaf path
因此递归返回的条件之一是
root.val == sum && root.left == null && root.right == null
左右子树有一个满足即可 (给子树的sum是每次传入的sum - root.val)
hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val)

Solution

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