Minimum Depth of Binary Tree -- LeetCode
2014-02-22 03:52
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原题链接:http://oj.leetcode.com/problems/minimum-depth-of-binary-tree/
这道题是树的题目,其实跟Maximum
Depth of Binary Tree非常类似,只是这道题因为是判断最小深度,所以必须增加一个叶子的判断(因为如果一个节点如果只有左子树或者右子树,我们不能取它左右子树中小的作为深度,因为那样会是0,我们只有在叶子节点才能判断深度,而在求最大深度的时候,因为一定会取大的那个,所以不会有这个问题)。这道题同样是递归和非递归的解法,递归解法比较常规的思路,比Maximum
Depth of Binary Tree多加一个左右子树的判断,代码如下:
这道题是树的题目,其实跟Maximum
Depth of Binary Tree非常类似,只是这道题因为是判断最小深度,所以必须增加一个叶子的判断(因为如果一个节点如果只有左子树或者右子树,我们不能取它左右子树中小的作为深度,因为那样会是0,我们只有在叶子节点才能判断深度,而在求最大深度的时候,因为一定会取大的那个,所以不会有这个问题)。这道题同样是递归和非递归的解法,递归解法比较常规的思路,比Maximum
Depth of Binary Tree多加一个左右子树的判断,代码如下:
public int minDepth(TreeNode root) { if(root == null) return 0; if(root.left == null) return minDepth(root.right)+1; if(root.right == null) return minDepth(root.left)+1; return Math.min(minDepth(root.left),minDepth(root.right))+1; }非递归解法同样采用层序遍历(相当于图的BFS),只是在检测到第一个叶子的时候就可以返回了,代码如下:
public int minDepth(TreeNode root) { if(root == null) return 0; LinkedList queue = new LinkedList(); int curNum = 0; int lastNum = 1; int level = 1; queue.offer(root); while(!queue.isEmpty()) { TreeNode cur = queue.poll(); if(cur.left==null && cur.right==null) return level; lastNum--; if(cur.left!=null) { queue.offer(cur.left); curNum++; } if(cur.right!=null) { queue.offer(cur.right); curNum++; } if(lastNum==0) { lastNum = curNum; curNum = 0; level++; } } return 0; }
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