剑指Offer 55 二叉树的深度

输入一棵二叉树，求该树的深度。从根结点到叶结点依次经过的结点（含根、叶结点）形成树的一条路径，最长路径的长度为树的深度。

 1 # -*- coding:utf-8 -*-
 2 # class TreeNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.left = None
 6 #         self.right = None
 7 class Solution:
 8     def __init__(self):
 9         self.maxlen = 0
10     
11     def preOrder(self,root,length):
12         if root != None:
13             length += 1
14             self.preOrder(root.left,length)
15             self.preOrder(root.right,length)
16             if root.left == None and root.right == None:
17                 self.maxlen = max(self.maxlen,length)
18             length -= 1
19         
20     def TreeDepth(self, pRoot):
21         if pRoot == None:
22             return 0
23         self.preOrder(pRoot,0)
24         return self.maxlen
25         # write code here

Java版代码，leetcode地址：

 1 class Solution {
 2     int maxLen = 0;
 3 
 4     public void preOrder(TreeNode root, int length) {
 5         if (root != null) {
 6             length++;// 遍历下一层节点，层数+1
 7             preOrder(root.left, length);
 8             preOrder(root.right, length);
 9             if (root.left == null && root.right == null) {
10                 // 叶子节点，计算一次最大深度
11                 maxLen = Integer.max(maxLen, length);
12             }
13             length--;// 遍历完叶子节点，回退，层数-1
14         }
15     }
16 
17     public int maxDepth(TreeNode root) {
18         if (root == null) {
19             return 0;
20         }
21         preOrder(root, 0);
22         return maxLen;
23     }
24 }

平衡二叉树

输入一棵二叉树，判断该二叉树是否是平衡二叉树。

 1 # -*- coding:utf-8 -*-
 2 # class TreeNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.left = None
 6 #         self.right = None
 7 class Solution:
 8     def __init__(self):
 9         self.isBalanced = True
10         
11     def height(self,root):
12         if root == None or not self.isBalanced:
13             return 0
14         left = self.height(root.left)
15         right = self.height(root.right)
16         if abs(left - right) > 1:
17             self.isBalanced = False
18         return 1 + max(left,right)
19         
20     def IsBalanced_Solution(self, pRoot):
21         self.height(pRoot)
22         return self.isBalanced
23         # write code here

Java版代码，leetcode地址：

 1 class Solution {
 2     boolean isBalanced = true;
 3 
 4     public int height(TreeNode root) {
 5         if (root == null || !isBalanced) {
 6             return 0;
 7         }
 8         int left = height(root.left);
 9         int right = height(root.right);
10         if (Math.abs(left - right) > 1) {
11             isBalanced = false;
12         }
13         return 1 + Integer.max(left, right);
14     }
15 
16     public boolean isBalanced(TreeNode root) {
17         height(root);
18         return isBalanced;
19     }
20 }

posted on 2019-06-14 16:02 Sempron2800+ 阅读(108) 评论(0) 编辑收藏举报