"Tree" Judge balanced binary tree AVL

1 /***********************************2 https://leetcode.com/problems/balanced-binary-tree/3 @date 2015.5.84 @description5 given a binary tree, determine whether it is a balanced binary tree6 @tags tree, DFS7 8 equilibrium binary tree satisfies the condition:9 1. The height difference of the left and right sub-tree is not greater than 1Ten 2. The left and right sub-trees are balanced binary trees One  A This solution offers two methods: - Solution1. - recursion first calculates the height of the left and right sub-trees, then compares the difference with 1, if ≤1, then judge whether the left and right sub-tree is AVL the  - Solution2. - calculates and returns the height of the left and right subtree, as well as determining whether the left and right subtrees are AVL trees, - this does not need to judge the left and right word height difference is not greater than 1 and then go back to determine whether the left and right sub-tree is AVL tree +  - added search binary tree creation and recursive first order traversal binary tree in test function +  A  at ************************************/ -  -  -#include <iostream> -#include <stdlib.h> -  in using namespacestd; -  to structtreenode{ +     intVal; -TreeNode *Left ; theTreeNode *Right ; *TreeNode (intx): Val (x), left (null), right (null) {} $ };Panax Notoginseng  -  the classsolution{ +  Public: A     BOOLisbalanced (TreeNode *root) { the         //Recursive +         if(!root)return true;//empty trees are also AVL -         intDepthleft = MaxDepth (root->Left ); $         intDepthright = MaxDepth (root->right);//The maximum depth of the left and right sub-tree, i.e. the height of the tree $         if(ABS (Depthleft-depthright) <=1)//the height difference of the left and right sub-tree is not more than 1, one condition -             returnisbalanced (Root->left) && isbalanced (root->right);//The left and right sub-trees are AVL, condition two -         Else the             return false; -     }Wuyi  the     intMaxDepth (TreeNode *root) { -         if(!root)return 0; Wu         return 1+ Max (maxDepth (Root->left), maxDepth (root->Right )); -     } About  $ }; -  -  - classsolution2{ A  Public: +     BOOLisbalanced (TreeNode *root) { the         returnDfsdepth (root)! =-1; -     } $  the     intDfsdepth (TreeNode *root) { the         if(!root)return 0; the         intLeftdepth = Dfsdepth (root->Left ); the         if(Leftdepth = =-1)return-1; -         intRightdepth = Dfsdepth (root->Right ); in         if(Rightdepth = =-1)return-1; the  the         if(ABS (Leftdepth-rightdepth) >1) About             return-1; the         returnMax (leftdepth, rightdepth) +1; the     } the }; +  -TreeNode *insert (TreeNode *root,intdata) { theTreeNode *ptr =Root;BayiTreeNode *tempnode;//The parent node of the inserted node is stored theTreeNode *newnode =NewTreeNode (data); the  -     if(ptr = =NULL) -         returnNewNode; the     Else{ the          while(PTR! =NULL) { theTempnode =ptr; the             if(Ptr->val >=data) { -PTR = ptr->Left ; the}Else{ thePTR = ptr->Right ; the             }94         } the         if(Tempnode->val >=data) { theTempnode->left =NewNode; the}Else{98Tempnode->right =NewNode; About         } -     }101     returnRoot;102 }103 104 //Recursive first-order traversal of binary tree the voidTravpre (TreeNode *root) {106     if(!root)return;107cout << Root->val <<" ";108Travpre (root->Left );109Travpre (root->Right ); the }111  the 113 intMain () { theTreeNode *root =NULL; the     inttemp =0; theCIN >>temp;117      while(Temp! =0){//End with 0 (input 0 terminates)118Root =Insert (root, temp);119CIN >>temp; -}//create a two-pronged tree121 122     //Recursive first-order traversal123 Travpre (root);124  thecout <<Endl;126 Solution2 A;127cout <<a.isbalanced (root); -}

"Tree" Judge balanced binary tree AVL