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.
Solution：Top-down
- 每次更新
newSum = sum - currentValue，传入左右子节点的recursion函数。 - 如果是
leaf node，且返回root.val == newSum??? - 如果不是
leaf node，则迭代得到左右子树是否能找到newSum，任一找到，就是true
return hasPathSum (root.left, newSum) || hasPathSum (root.right, newSum);
/**
* 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.left == null && root.right == null) {
return root.val == sum;
}
int newSum = sum - root.val;
return hasPathSum (root.left, newSum) || hasPathSum (root.right, newSum);
}
}
