112.Path Sum (数的路径为N)

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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:
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.

public 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) ;
}

return (hasPathSum(root.left, sum-root.val)||hasPathSum(root.right, sum-root.val));

}
}

------------------------------------------------------------------

void helper(TreeNode * root, int sum, int path[], int top) {
path[top++] = root.data;
sum -= root.data;
if (root->left == NULL && root->right==NULL) {
    if (sum == 0) printPath(path, top);
} else {
    if (root->left != NULL) helper(root->left, sum, path, top);
    if (root->right!=NULL) helper(root->right, sum, path, top);
}
top --;
sum += root.data; //....
}

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