二叉树中和为某一值的路径算法问题

[yuwengCipher]

function FindPathCore(node, expectNumber, stack, sum, result) {
stack.push(node.val);
sum += node.val;
if (!node.left && !node.right && sum === expectNumber) {
result.push(stack.slice(0));
}
if (node.left) {
FindPathCore(node.left, expectNumber, stack, sum, result);
}
if (node.right) {
FindPathCore(node.right, expectNumber, stack, sum, result);
}
stack.pop();
}

当stack[stack.length-1]存在并且没访问过时，stack.pop()会将右子树切断，造成结果项缺失。