在 BST 中查找节点位置并将其递归添加到树中
finding a node location within a BST and adding it to the tree recursively
所以我有 3 种方法 1 使用 traverseAdd 方法将节点添加到二叉树,另一种方法根据其父节点查找值在树中的放置位置。我想消除 traverseAdd 方法并在 add 方法中使用 findLocation 方法将新值添加到 BST。
public void add(int val) {
/*Adds a new node to the binary tree after traversing the tree and figuring out where it belongs*/
Node nodeObjToAdd = new Node(val);
if(root == null){
//if node root is not null root = new node value
root = nodeObjToAdd;
}
Node nodeTraversed = root;
traverseAdd(nodeTraversed, nodeObjToAdd);
}
private void traverseAdd(Node node, Node nodeObjToAdd){
/*Traverses tree and finds a place to add the node to be added by comparing values of the left child and right child of the
* focus node*/
if(nodeObjToAdd.value < node.value){
if(node.leftChild == null){
node.leftChild = nodeObjToAdd;
}
else {
//if the val < the root.value set int he constructor
traverseAdd(node.leftChild, nodeObjToAdd);
}
}
else if(nodeObjToAdd.value > node.value) {
if (node.rightChild == null) {
node.rightChild = nodeObjToAdd;
} else {
traverseAdd(node.rightChild, nodeObjToAdd);
}
}
}
public Node findNodeLocation(Node rootNode, int val) {
/*returns where a the Node after which the value would be added.*/
if(val < rootNode.value && rootNode.leftChild != null){
return rootNode.leftChild;
}
if(val >= rootNode.value && rootNode.rightChild != null){
return rootNode.rightChild;
}
else
return this.root;
}
public void add(int val) {
if (root == null) {
root = new Node(val);
}
Node cur = root;
Node next = null;
while (true) {
next = findNodeLocation(cur, val);
if (next != cur) {
cur = next;
} else {
break;
}
}
if (val < cur.value) {
cur.leftChild = new Node(val);
} else {
cur.rightChild = new Node(val);
}
}
我认为这应该可行
所以我有 3 种方法 1 使用 traverseAdd 方法将节点添加到二叉树,另一种方法根据其父节点查找值在树中的放置位置。我想消除 traverseAdd 方法并在 add 方法中使用 findLocation 方法将新值添加到 BST。
public void add(int val) {
/*Adds a new node to the binary tree after traversing the tree and figuring out where it belongs*/
Node nodeObjToAdd = new Node(val);
if(root == null){
//if node root is not null root = new node value
root = nodeObjToAdd;
}
Node nodeTraversed = root;
traverseAdd(nodeTraversed, nodeObjToAdd);
}
private void traverseAdd(Node node, Node nodeObjToAdd){
/*Traverses tree and finds a place to add the node to be added by comparing values of the left child and right child of the
* focus node*/
if(nodeObjToAdd.value < node.value){
if(node.leftChild == null){
node.leftChild = nodeObjToAdd;
}
else {
//if the val < the root.value set int he constructor
traverseAdd(node.leftChild, nodeObjToAdd);
}
}
else if(nodeObjToAdd.value > node.value) {
if (node.rightChild == null) {
node.rightChild = nodeObjToAdd;
} else {
traverseAdd(node.rightChild, nodeObjToAdd);
}
}
}
public Node findNodeLocation(Node rootNode, int val) {
/*returns where a the Node after which the value would be added.*/
if(val < rootNode.value && rootNode.leftChild != null){
return rootNode.leftChild;
}
if(val >= rootNode.value && rootNode.rightChild != null){
return rootNode.rightChild;
}
else
return this.root;
}
public void add(int val) {
if (root == null) {
root = new Node(val);
}
Node cur = root;
Node next = null;
while (true) {
next = findNodeLocation(cur, val);
if (next != cur) {
cur = next;
} else {
break;
}
}
if (val < cur.value) {
cur.leftChild = new Node(val);
} else {
cur.rightChild = new Node(val);
}
}
我认为这应该可行