此方法采用整数树并构造以下字符串:
(根部的数据,左子树的String,右子树的String)
例如,如果变量树存储对以下树的引用:
+---+
| 2 |
+---+
/ \
+---+ +---+
| 8 | | 1 |
+---+ +---+
/ / \
+---+ +---+ +---+
| 0 | | 7 | | 6 |
+---+ +---+ +---+
/ \
+---+ +---+
| 4 | | 9 |
+---+ +---+
它应该返回一个字符串:
"(2, (8, 0, empty), (1, (7, 4, empty), (6, empty, 9)))"
对于空树,该方法应返回“empty”。对于叶节点,它应该以字符串形式返回节点中的数据。对于分支节点,它应该返回一个带括号的字符串,该字符串包含三个以逗号分隔的元素:
使用我的方法,它有时会起作用,但有时会生成 NullPointerException。谁能指出发生这种情况的地点和原因?
这是我的:
private IntTreeNode overallRoot; // first node; linked to other nodes
// post: returns the string of all data in leaves. For a branch node, return a parenthesized String
// w/ three elementsseparated by commas. return empty for empty tree
public String toString2() {
if (overallRoot == null) { // If empty tree
return "empty";
} else {
return toString2(overallRoot);
}
}
// helper for toString
private String toString2(IntTreeNode root) {
String value = "";
if (root.left != null && root.right != null) { // If both branches exist
value += "(" + root.data + ", " + toString2(root.left)+ ", " + toString2(root.right) + ")";
} else if (root.left != null && root.right == null) { // if right branch is empty
value += "(" + root.data + ", " + toString2(root.left) + ", empty)";
} else if (root.left == null && root.right != null) { // if left branch is empty
value += "(" + root.data + ", empty, " + toString2(root.left) + ")";
} else { // If at a leaf
return "" + root.data;
}
return value;
}
这是节点类:
public class IntTreeNode {
public int data;
public IntTreeNode left;
public IntTreeNode right;
// constructs a leaf node with given data
public IntTreeNode(int data) {
this(data, null, null);
}
// constructs a branch node with given data, left subtree,
// right subtree
public IntTreeNode(int data, IntTreeNode left, IntTreeNode right) {
this.data = data;
this.left = left;
this.right = right;
}
}
最佳答案
错误在这里:
else if (root.left == null && root.right != null) { // if left branch is empty
value += "(" + root.data + ", empty, " + toString2(root.left) + ")";
}
应该是……
else if (root.left == null && root.right != null) { // if left branch is empty
value += "(" + root.data + ", empty, " + toString2(root.right) + ")";
}
...,将 root.right
而不是 root.left
传递给 toString2
。
关于java - 二叉树和 NullPointerException,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25151268/