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.
Tips：在一个给定的树上找出是否包含和为sum的路径（路径指从root到根节点）
我采用的递归方法，root有左子树，就用root.left调用作为参数，再次调用函数。
root有右子树，就用root.right调用作为参数，再次调用函数。
public boolean hasPathSum(TreeNode root, int sum) { if (root == null) return false; boolean has = false; int ans = 0; has = hasPathSumCore(root, sum, ans); return has; } private boolean hasPathSumCore(TreeNode root, int sum, int ans) { // TODO Auto-generated method stub boolean has = false; if (root == null) { System.out.println("null"); return false; } ans += root.val; System.out.println(ans); if (ans == sum && root.left == null && root.right == null) { has = true; } if (root.left != null) { boolean has1 = hasPathSumCore(root.left, sum, ans); has = has || has1; } if (root.right != null) { boolean has2 = hasPathSumCore(root.right, sum, ans); has = has || has2; } ans -= root.val; return has; }
之后在leetcode上看到更简单的做法（-_-||）
public boolean hasPathSum2(TreeNode root, int sum) { if(root == null) return false; if(root.left == null && root.right == null && sum - root.val == 0) return true; return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val); }
