Maximum Depth of Binary Tree -- LeetCode
2014-02-22 03:34
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原题链接:http://oj.leetcode.com/problems/maximum-depth-of-binary-tree/
这是一道比较简单的树的题目,可以有递归和非递归的解法,递归思路简单,返回左子树或者右子树中大的深度加1,作为自己的深度即可,代码如下:
Depth of Binary Tree.
这是一道比较简单的树的题目,可以有递归和非递归的解法,递归思路简单,返回左子树或者右子树中大的深度加1,作为自己的深度即可,代码如下:
public int maxDepth(TreeNode root) { if(root == null) return 0; return Math.max(maxDepth(root.left),maxDepth(root.right))+1; }非递归解法一般采用层序遍历(相当于图的BFS),因为如果使用其他遍历方式也需要同样的复杂度O(n). 层序遍历理解上直观一些,维护到最后的level便是树的深度。代码如下:
public int maxDepth(TreeNode root) { if(root == null) return 0; int level = 0; LinkedList<TreeNode> queue = new LinkedList<TreeNode>(); queue.add(root); int curNum = 1; //num of nodes left in current level int nextNum = 0; //num of nodes in next level while(!queue.isEmpty()) { TreeNode n = queue.poll(); curNum--; if(n.left!=null) { queue.add(n.left); nextNum++; } if(n.right!=null) { queue.add(n.right); nextNum++; } if(curNum == 0) { curNum = nextNum; nextNum = 0; level++; } } return level; }总体来说我觉得这道题可以考核树的数据结构,也可以看看对递归和非递归的理解。相关的扩展可以是Minimum
Depth of Binary Tree.
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