我目前正在编写一个方法,在Java中搜索以最短距离到达两个节点的节点。我的想法是,如果树中都存在两个节点,则根将是第一个可以同时到达的节点。所以我递归并检查根的左边/右边,看看它们是否也能达到两者。通过找到至少找到一个节点后找不到的第一个节点,我应该找到距离我所搜索的节点最近的节点。Java二叉树:找到达到两个节点的距离最短的节点
我已经将此任务分解为两个方法,一个名为canReach,它搜索树中的一个节点,另一个使用canReach的布尔返回来确定向下移动树,并沿着什么方向命名为reachingBoth。
通过调试我非常有信心canReach准确地搜索树。然而,如果两个节点都不存在于树中,或者如果它们是根,那么从来没有任何介于两者之间的东西。
重复检查两个节点的访问,同时向下导航树是一个好主意?如果任何人都可以看到我的方法到达哪里都有问题,我会很感激这种洞察力。
public boolean canReach(T d) {
// Helper method for reachesBoth. Takes an argument of data of a Node
int comp = d.compareTo(data); // Compare data to current node
if (comp == 0) return true; // Found the node
if (comp < 0) { // search left for the node
if (left != null) {
if (left.canReach(d) == true) return true;
}
return false;
}
if (comp > 0) { // search right for the node
if (right != null) {
if (right.canReach(d) == true) return true;
}
return false;
}
return false; // Cannot find the node in our tree
}
public T reachesBoth(T a, T b) {
if (canReach(a) == true && canReach(b) == true) {
// Found the first node that can reach both desired nodes.
// Must continue to see if a node with shorter distance
// can reach both nodes as well.
if (left != null && right != null) {
// Case 1: Two more branches to check
if (left.reachesBoth(a, b) != null) left.reachesBoth(a, b);
if (right.reachesBoth(a, b) != null) right.reachesBoth(a, b);
//Both branches cannot reach both nodes sooner than we can.
if (left.reachesBoth(a, b) == null & right.reachesBoth(a, b) == null) {
return this.data;
}
}
if (left != null && right == null) {
// Case 2: Only left branch to check
if (left.reachesBoth(a, b) == null) return this.data;
}
if (left == null && right != null) {
// Case 3: Only right branch to check
if (right.reachesBoth(a, b) == null) return this.data;
}
// Case 4: No more tree to search for a better solution
if (left == null && right == null) return this.data;
}
return null; // Cannot locate both nodes in our tree from our start
}
这奏效了,谢谢! –