1,把二叉树的所有节点值遍历并添加到数组
public void addList(TreeNode node) {
if (node != null) {
list.add(node.val);
if (node.left != null)
addList(node.left);
if (node.right != null)
addList(node.right);
}
}
各种遍历
* 先序遍历二叉树(递归)
* @param node
*/
public void preOrderTraverse(Node<E> node) {
System.out.print(node.value + " ");
if (node.left != null)
preOrderTraverse(node.left);
if (node.right != null)
preOrderTraverse(node.right);
}
/**
* 中序遍历二叉树(递归)
* @param node
*/
public void inOrderTraverse(Node<E> node) {
if (node.left != null)
inOrderTraverse(node.left);
System.out.print(node.value + " ");
if (node.right != null)
inOrderTraverse(node.right);
}
/**
* 后序遍历二叉树(递归)
* @param node
*/
public void postOrderTraverse(Node<E> node) {
if (node.left != null)
postOrderTraverse(node.left);
if (node.right != null)
postOrderTraverse(node.right);
System.out.print(node.value + " ");
}
/**
* 先序遍历二叉树(非递归)
* @param root
*/
public void preOrderTraverseNoRecursion(Node<E> root) {
LinkedList<Node<E>> stack = new LinkedList<Node<E>>();
Node<E> currentNode = null;
stack.push(root);
while (!stack.isEmpty()) {
currentNode = stack.pop();
System.out.print(currentNode.value + " ");
if (currentNode.right != null)
stack.push(currentNode.right);
if (currentNode.left != null)
stack.push(currentNode.left);
}
}
/**
* 中序遍历二叉树(非递归)
* @param root
*/
public void inOrderTraverseNoRecursion(Node<E> root) {
LinkedList<Node<E>> stack = new LinkedList<Node<E>>();
Node<E> currentNode = root;
while (currentNode != null || !stack.isEmpty()) {
// 一直循环到二叉排序树最左端的叶子结点(currentNode是null)
while (currentNode != null) {
stack.push(currentNode);
currentNode = currentNode.left;
}
currentNode = stack.pop();
System.out.print(currentNode.value + " ");
currentNode = currentNode.right;
}
}
/**
* 后序遍历二叉树(非递归)
* @param root
*/
public void postOrderTraverseNoRecursion(Node<E> root) {
LinkedList<Node<E>> stack = new LinkedList<Node<E>>();
Node<E> currentNode = root;
Node<E> rightNode = null;
while (currentNode != null || !stack.isEmpty()) {
// 一直循环到二叉排序树最左端的叶子结点(currentNode是null)
while (currentNode != null) {
stack.push(currentNode);
currentNode = currentNode.left;
}
currentNode = stack.pop();
// 当前结点没有右结点或上一个结点(已经输出的结点)是当前结点的右结点,则输出当前结点
while (currentNode.right == null || currentNode.right == rightNode) {
System.out.print(currentNode.value + " ");
rightNode = currentNode;
if (stack.isEmpty()) {
return; //root以输出,则遍历结束
}
currentNode = stack.pop();
}
stack.push(currentNode); // 还有右结点没有遍历
currentNode = currentNode.right;
}
}
/**
* 广度优先遍历二叉树,又称层次遍历二叉树
* @param node
*/
public void breadthFirstTraverse(Node<E> root) {
Queue<Node<E>> queue = new LinkedList<Node<E>>();
Node<E> currentNode = null;
queue.offer(root);
while (!queue.isEmpty()) {
currentNode = queue.poll();
System.out.print(currentNode.value + " ");
if (currentNode.left != null)
queue.offer(currentNode.left);
if (currentNode.right != null)
queue.offer(currentNode.right);
}
}
深度遍历计算二叉树的深度
public int maxDepth(Node root) {
if (root == null) return 0;
int depth = 1;
for (Node children : root.children)
depth = Math.max(depth, maxDepth(children) + 1);
return depth;
}
非递归实现二叉树三种遍历方式
先序遍历:
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<>();
if(root == null) return list;
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while(!stack.empty()){
root = stack.pop();
list.add(0, root.val);
if(root.left != null) stack.push(root.left);
if(root.right != null) stack.push(root.right);
}
return list;
}
后续遍历:
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<>();
if(root == null) return list;
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while(!stack.empty()){
root = stack.pop();
list.add(root.val);
if(root.right != null) stack.push(root.right);
if(root.left != null) stack.push(root.left);
}
return list;
}
中序遍历:
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<>();
if(root == null) return list;
Stack<TreeNode> stack = new Stack<>();
while(root != null || !stack.empty()){
while(root != null){
stack.push(root);
root = root.left;
}
root = stack.pop();
list.add(root.val);
root = root.right;
}
return list;
}