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判断树中结点个数    public static <T> int getTreeNum(TreeNode<T> root)

判断树中结点个数 public static int getTreeNum(TreeNode root)

return getTreeNum(root.leftChild) + getTreeNum(root.rightChild) + 1;

判断树的深度 public static int getTreeDepth(TreeNode root)

		int leftDepth = getTreeDepth(root.leftChild) + 1;
        int rightDepth = getTreeDepth(root.rightChild) + 1;
        return Math.max(leftDepth, rightDepth);

前序遍历 public static void preOrderTravel(TreeNode root)

		if (root == null) {
            return;
        }
        visitNode(root);
        preOrderTravel(root.leftChild);
        preOrderTravel(root.rightChild);

中序遍历 public static void midOrderTravel(TreeNode root)

		if (root == null) {
            return;
        }
        midOrderTravel(root.leftChild);
        visitNode(root);
        midOrderTravel(root.rightChild);

后序遍历 public static void backOrderTravel(TreeNode root)

    	if (root == null) {
            return;
        }
        midOrderTravel(root.leftChild);
        
        midOrderTravel(root.rightChild);
        visitNode(root);

层次遍历 public static void levelTravel(TreeNode root)

		Queue<TreeNode<T>> q = new LinkedList();
        q.offer(root);  //入队列

        while (!q.isEmpty()) {

            TreeNode<T> temp = q.poll();    //出队列
            visitNode(temp);
            if (temp.leftChild != null) {
                q.offer(temp.leftChild);
            }
            if (temp.rightChild != null) {
                q.offer(temp.rightChild);
            }

        }

访问node结点 private static void visitNode

		System.out.print(node.value + "\t");

求第K层结点个数 public static int getNumForKlevel(TreeNode root, int k)

		int leftNum = getNumForKlevel(root.leftChild, k - 1);
        int rightNum = getNumForKlevel(root.rightChild, k - 1);
        
        return leftNum + rightNum;

根据前序和中序构建二叉树（有点难度哦！）

	//TreeNode<Integer> root = TreeTools.getTreeFromPreAndMid(pre, mid);
    //pre = [1, 2, 4, 5, 3]           mid = [4, 2, 5, 1, 3]
    public static <T> TreeNode<T> getTreeFromPreAndMid(List<T> pre, List<T> mid) {
        if (pre == null || mid == null || pre.size() == 0 || mid.size() == 0) {
            return null;
        }
        if (pre.size() == 1) {
            return new TreeNode<T>(pre.get(0));
        }
        TreeNode<T> root = new TreeNode<T>(pre.get(0));
        // 找出根节点在中序中的位置
        int index = 0;
        while (!mid.get(index++).equals(pre.get(0))) {
        }


        // 构建左子树的前序
        List<T> preLeft = new ArrayList<T>(index);  //index = 3

        // 左子树的中序
        List<T> midLeft = new ArrayList<T>(index);
        
        for (int i = 1; i < index; i++) {
            preLeft.add(pre.get(i));
        }
        for (int i = 0; i < index - 1; i++) {
            midLeft.add(mid.get(i));
        }

        // 重建左子树
        root.leftChild = getTreeFromPreAndMid(preLeft, midLeft);
        
        
        // 右子树的前序
        List<T> preRight = new ArrayList<T>(pre.size() - index - 1);
        // 右子树的中序
        List<T> midRight = new ArrayList<T>(pre.size() - index - 1);
        for (int i = 0; i <= pre.size() - index - 1; i++) {
            preRight.add(pre.get(index + i));
        }
        for (int i = 0; i <= pre.size() - index - 1; i++) {
            midRight.add(mid.get(index + i));
        }
        // 重建→子树
        root.rightChild = getTreeFromPreAndMid(preRight, midRight);
        return root;
    }

原文链接:https://blog.csdn.net/AC_872767407/article/details/114415353