[剑指Offer] 38.二叉树的深度

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。

【思路1】递归

 1 /*
 2 struct TreeNode {
 3     int val;
 4     struct TreeNode *left;
 5     struct TreeNode *right;
 6     TreeNode(int x) :
 7             val(x), left(NULL), right(NULL) {
 8     }
 9 };*/
10 class Solution {
11 public:
12     int TreeDepth(TreeNode* pRoot)
13     {
14         if(pRoot == NULL)
15             return 0;
16         int h1 = TreeDepth(pRoot->left);
17         int h2 = TreeDepth(pRoot->right);
18         return max(h1,h2) + 1;
19     }
20 };

【思路2】DFS，用一个栈来存储结点，一个栈来存储当前深度

 1 /*
 2 struct TreeNode {
 3     int val;
 4     struct TreeNode *left;
 5     struct TreeNode *right;
 6     TreeNode(int x) :
 7             val(x), left(NULL), right(NULL) {
 8     }
 9 };*/
10 class Solution
11 {
12 public:
13     int TreeDepth(TreeNode* pRoot)
14     {
15         if(pRoot == NULL) return 0;
16         stack<TreeNode*> nodeStack;
17         stack<int> depthStack;
18         
19         nodeStack.push(pRoot);
20         depthStack.push(1);
21         
22         TreeNode* node = pRoot;
23         int MaxDepth = 0;
24         int curDepth = 0;
25         
26         while(!nodeStack.empty())
27         {
28             node = nodeStack.top();
29             nodeStack.pop();
30             curDepth = depthStack.top();
31             depthStack.pop();
32             if(MaxDepth < curDepth)
33                 MaxDepth = curDepth;
34             
35             if(node->left != NULL)
36             {
37                 nodeStack.push(node->left);
38                 depthStack.push(curDepth + 1);
39             }
40             if(node->right != NULL)
41             {
42                 nodeStack.push(node->right);
43                 depthStack.push(curDepth + 1);
44             }
45         }
46         return MaxDepth;
47     }
48 };