05 August 2008
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’,
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7
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.
典型的递归解法，每次递归以后注意sum需要减去当前root的val。
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/**
* 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) {
return dfs(root, sum);
}
private boolean dfs(TreeNode root, int sum) {
if (root == null) {
return false;
}
if (root.left == null && root.right == null && sum == root.val) {
return true;
}
return dfs(root.left, sum - root.val) || dfs(root.right, sum - root.val);
}
}