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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: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.
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显然深搜求解，简单高效。注意叶子节点的定义，当且仅当左右孩子都为空时，才是叶子节点。如果一个节点有一个孩子时，是不能作为leaf的。
我的AC代码
public class PathSum {
static Boolean ok = false;
/**
* @param args
*/
public static void main(String[] args) {
TreeNode treeNode = new TreeNode(1);
TreeNode t1 = new TreeNode(2);
System.out.println(hasPathSum(treeNode, 1));
System.out.println(hasPathSum(null, 0));
treeNode.left = t1;
System.out.println(hasPathSum(treeNode, 1));
}
public static boolean hasPathSum(TreeNode root, int sum) {
if(root == null) return false;
ok = false;
dfs(root, sum, 0);
return ok;
}
public static void dfs(TreeNode root, int sum, int s) {
if(ok == true) return;
if(root.left == null && root.right == null) {
if (s + root.val == sum) {
ok = true;
}
return;
}
if(root.left != null) dfs(root.left, sum, s + root.val);
if(root.right != null) dfs(root.right, sum, s + root.val);
}
}
LeetCode Oj 112. Path Sum 解题报告
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原文地址：http://blog.csdn.net/bruce128/article/details/50716386
