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			74 lines
		
	
	
		
			2.1 KiB
		
	
	
	
		
			C#
		
	
	
	
	
	
			
		
		
	
	
			74 lines
		
	
	
		
			2.1 KiB
		
	
	
	
		
			C#
		
	
	
	
	
	
/**
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 * File: TreeNode.cs
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 * Created Time: 2022-12-23
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 * Author: haptear (haptear@hotmail.com)
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 */
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namespace hello_algo.utils;
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/* 二叉树节点类 */
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public class TreeNode {
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    public int val;        // 节点值
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    public int height;     // 节点高度
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    public TreeNode? left;  // 左子节点引用
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    public TreeNode? right; // 右子节点引用
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    public TreeNode(int x) {
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        val = x;
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    }
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    /* Generate a binary tree given an array */
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    public static TreeNode? ListToTree(List<int?> arr) {
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        if (arr.Count == 0 || arr[0] == null)
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            return null;
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        TreeNode root = new TreeNode(arr[0]!.Value);
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        Queue<TreeNode> queue = new Queue<TreeNode>();
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        queue.Enqueue(root);
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        int i = 0;
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        while (queue.Count != 0) {
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            TreeNode node = queue.Dequeue();
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            if (++i >= arr.Count) break;
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            if (arr[i] != null) {
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                node.left = new TreeNode((int)arr[i]);
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                queue.Enqueue(node.left);
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            }
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            if (++i >= arr.Count) break;
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            if (arr[i] != null) {
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                node.right = new TreeNode((int)arr[i]);
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                queue.Enqueue(node.right);
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            }
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        }
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        return root;
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    }
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    /* Serialize a binary tree to a list */
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    public static List<int?> TreeToList(TreeNode root) {
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        List<int?> list = new();
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        if (root == null) return list;
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        Queue<TreeNode?> queue = new();
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        while (queue.Count != 0) {
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            TreeNode? node = queue.Dequeue();
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            if (node != null) {
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                list.Add(node.val);
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                queue.Enqueue(node.left);
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                queue.Enqueue(node.right);
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            } else {
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                list.Add(null);
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            }
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        }
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        return list;
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    }
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    /* Get a tree node with specific value in a binary tree */
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    public static TreeNode? GetTreeNode(TreeNode? root, int val) {
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        if (root == null)
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            return null;
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        if (root.val == val)
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            return root;
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        TreeNode? left = GetTreeNode(root.left, val);
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        TreeNode? right = GetTreeNode(root.right, val);
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        return left ?? right;
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    }
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}
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