[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

 /*
 struct TreeNode {
     int val;
     struct TreeNode *left;
     struct TreeNode *right;
     TreeNode(int x) :
             val(x), left(NULL), right(NULL) {
     }
 };*/
 class Solution {
 public:
     int TreeDepth(TreeNode* pRoot)
     {
         if(pRoot == NULL)
             return ;
         int h1 = TreeDepth(pRoot->left);
         int h2 = TreeDepth(pRoot->right);
         return max(h1,h2) + ;
     }
 };

【思路2】DFS，用一个栈来存储结点，一个栈来存储当前深度

 /*
 struct TreeNode {
     int val;
     struct TreeNode *left;
     struct TreeNode *right;
     TreeNode(int x) :
             val(x), left(NULL), right(NULL) {
     }
 };*/
 class Solution
 {
 public:
     int TreeDepth(TreeNode* pRoot)
     {
         if(pRoot == NULL) return ;
         stack<TreeNode*> nodeStack;
         stack<int> depthStack;

         nodeStack.push(pRoot);
         depthStack.push();

         TreeNode* node = pRoot;
         int MaxDepth = ;
         int curDepth = ;

         while(!nodeStack.empty())
         {
             node = nodeStack.top();
             nodeStack.pop();
             curDepth = depthStack.top();
             depthStack.pop();
             if(MaxDepth < curDepth)
                 MaxDepth = curDepth;

             if(node->left != NULL)
             {
                 nodeStack.push(node->left);
                 depthStack.push(curDepth + );
             }
             if(node->right != NULL)
             {
                 nodeStack.push(node->right);
                 depthStack.push(curDepth + );
             }
         }
         return MaxDepth;
     }
 };