劍指offer 面試題55 二叉樹的深度&平衡二叉樹

二叉樹的深度：
使用遞歸，複雜度與樹的結點樹有關

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 0;
        }
        int left=TreeDepth(pRoot->left);
        int right=TreeDepth(pRoot->right);
        return left>right ? left+1: right+1;
    }
};

平衡二叉樹
思路與上題類似，這裏用一個引用變量來存儲樹的深度

#include <iostream>
#include <string>
#include <memory>
#include <vector>
#include<cmath>
#include<algorithm>
#include<stack>
#include<map>
#include<sstream>
using namespace std;


struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};

class Solution {
public:
    bool IsBalanced_Solution(TreeNode* pRoot) {
      int depth=0;
      return isBalancedTreeWithDepth(pRoot,depth);
    }
    bool isBalancedTreeWithDepth(TreeNode* pRoot,int & depth) {
       if(pRoot==NULL) {
          return true;
       }
       int leftDepth=0,rightDepth=0;
       bool leftRes=isBalancedTreeWithDepth(pRoot->left,leftDepth);
       bool rightRes=isBalancedTreeWithDepth(pRoot->right,rightDepth);
      depth=leftDepth>rightDepth ? leftDepth+1 : rightDepth+1;
      if(abs(leftDepth-rightDepth)>1) {
         return false;
      } else {
         return true;
      }
    }
    
};

int main()
{
   //  vector<int> test={3,1,2};
    Solution sol;
    TreeNode* test=new TreeNode(3);
   test->left=new TreeNode(1);
   test->left->left=new TreeNode(0);
   int res=sol.IsBalanced_Solution(test);
   cout<<res<<endl;
   system("pause");
    return 0;
}