题目描述:
输入一棵二叉树,求该树的深度。从根结点到叶结点依次通过的结点(含根、叶结点)造成树的一条路径,最长路径的长度为树的深度。java
题解:node
题解一:
递归实现。web
public int TreeDepth(TreeNode root) { if(root == null){ return 0; } int left = TreeDepth (root.left); int right = TreeDepth (root.right); return Math.max (left,right) + 1; }
题解二:
利用队列来解决。
在出队以前,此时队列中记录的只有某一层节点,因此队列的大小就是某一层节点的个数。当此个数减到 0 的时候,则说明该层节点所有出队完成。svg
public int TreeDepth2(TreeNode root) { if (root == null) { return 0; } Queue<TreeNode> queue = new LinkedList<TreeNode> (); queue.add (root); int depth = 0; int size = -1; while (!queue.isEmpty()) { size = queue.size (); while (size != 0) { TreeNode node = queue.poll (); if (node.left != null) { queue.add (node.left); } if (node.right != null) { queue.add (node.right); } size--; } depth++; } return depth; }
解法三:
层序遍历。须要一个list来存某一层的所有结点。spa
public int TreeDepth(TreeNode root) { if (root == null) { return 0; } ArrayList<TreeNode> list = new ArrayList<TreeNode> (); list.add (root); int depth = 0; while (list.size () != 0) { depth++; ArrayList<TreeNode> tmp = new ArrayList<> (); for (int i = 0; i < list.size (); i++) { while (list.get (i).left != null) { tmp.add (list.get (i).left); } while (list.get (i).right != null) { tmp.add (list.get (i).right); } } // list中存着一层结点 list = tmp; } return depth; }