Question

我有一个BST的分配，我必须找到最接近给定节点的节点。显然，如果节点在树中，它将被返回。否则，需要返回的节点是树上最近的节点（可以在两侧）。

脚本工作正常，但节点存在的情况除外。该函数不再返回节点，而是继续运行并吐出不同的节点而不是杀死该函数。我想知道是否有人可以帮助我。

树的结构如下：

它由节点
每个节点都有一个“条目”（一个字符串值），一个“右”和一个“左”

这就是我所拥有的，谢谢！

    public Entry getClosestEntry(String w) { 
            if (root == null) return null;
            else if (root.left == null && root.right == null) return root.entry;
            else return getClosestEntry(w, root, null, null, null); 
    }
    private Entry getClosestEntry(String w, Node baseNode, Node highestLow, Node lowestHi, Node finalClosest) {
        System.out.println("Dict: " + this);
        System.out.println("Top string to compare: " + w);
        System.out.println("Top baseNode: " + baseNode.entry.word);
        if (w.compareTo(baseNode.entry.word) == 0) {
            System.out.println("Top finalClosest is baseNode: " + baseNode.entry.word);
            finalClosest = baseNode;
        } else {
            if (highestLow != null) System.out.println("Top highestLow: " + highestLow.entry.word);
            else System.out.println("Top highestLow is null");
            if (lowestHi != null) System.out.println("Top lowestHi: " + lowestHi.entry.word);
            else System.out.println("Top lowestHi is null");
            if (finalClosest != null) System.out.println("Top finalClosest: " + finalClosest.entry.word);
            else System.out.println("Top finalClosest is null");

            int cmp = w.compareTo(baseNode.entry.word);

            if (cmp < 0) {
                System.out.println("Word is less than base.");
                if (lowestHi == null) {
                    lowestHi = baseNode;
                    System.out.println("lowestHi set to: " + lowestHi.entry.word);
                }
                else {
                    if (w.compareTo(lowestHi.entry.word) < 0 && baseNode.entry.word.compareTo(lowestHi.entry.word) < 0) {
                        lowestHi = baseNode;
                        System.out.println("lowestHi changed to: " + lowestHi.entry.word);
                    }
                }
                System.out.println("Returning right side of base.");
                if (baseNode.right != null) getClosestEntry(w, baseNode.right, highestLow, lowestHi, finalClosest);
            } else { 
                System.out.println("Word is greater than base.");
                if (highestLow == null) {
                    highestLow = baseNode;
                    System.out.println("highestLow set to: " + highestLow.entry.word);
                }
                else {
                    if (w.compareTo(highestLow.entry.word) > 0 && baseNode.entry.word.compareTo(highestLow.entry.word) > 0) {
                        highestLow = baseNode;
                        System.out.println("highestLow changed to: " + highestLow.entry.word);
                    }
                }
                System.out.println("Returning left side of base.");
                if (baseNode.left != null) getClosestEntry(w, baseNode.left, highestLow, lowestHi, finalClosest);
            }

            if (lowestHi == null && highestLow != null && finalClosest == null) {
                System.out.println("lowestHi is null, so finalClosest must be highestLow.");
                finalClosest = highestLow;
            } else if (highestLow == null && lowestHi != null && finalClosest == null) {
                System.out.println("highestLow is null, so finalClosest must be lowestHi.");
                finalClosest = lowestHi;
            } else if (lowestHi == null && highestLow == null && finalClosest == null) {
                System.out.println("Both sides are null, so node must be null.");
                return null;
            } else {
                System.out.println("Both sides are there. Default to highestLow if finalClosest is null");
                if (finalClosest == null) finalClosest = highestLow;
            }
        }
        System.out.println("Final Closest: " + finalClosest.entry.word);
        return finalClosest.entry;
    }

Answer 1

修正了它。我以前忘了解决某些情况。

返回语句后Java函数继续

1 个答案: