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 andsum = 22,5 / 4 8 / / 11 13 4 / \ 7 2 1return true, as there exist a root-to-leaf path
5->4->11->2which sum is 22.
本质是树的遍历，使用树的先序遍历，并使用一个变量tempSum来记录遍历到当前结点的和。
public boolean preOrder(TreeNode node, int sum, int tempSum) { if (node != null) { tempSum += node.val; if (node.left == null && node.right == null) { if (tempSum == sum) { return true; } else { return false; } } return preOrder(node.left, sum, tempSum) || preOrder(node.right, sum, tempSum); } return false; } public boolean hasPathSum(TreeNode root, int sum) { return preOrder(root, sum, 0); }
原文地址：http://blog.csdn.net/u010378705/article/details/28431253
