递归求解:
递归结束条件:节点为空,则其深度为0
否则以root为根节点的子树的深度等于1加上左右子树的深度大的那一个
public int TreeDepth(TreeNode root) { if (root == null) { return 0; } return 1 + Math.max(TreeDepth(root.left), TreeDepth(root.right)); }
迭代求解:层序遍历
每当遍历完一层,深度加一
import java.util.LinkedList; import java.util.Queue; public class Solution { public int TreeDepth(TreeNode root) { if (root == null) { return 0; } Queue<TreeNode> queue = new LinkedList<>(); int count = 0; queue.add(root); while (!queue.isEmpty()) { int size = queue.size(); for (int i = 0; i < queue.size(); i++) { TreeNode tem = queue.poll(); if (tem.left != null) { queue.add(tem.left); } if (tem.right != null) { queue.add(tem.right); } } count++; } return count; } }