前序遍歷重構二叉樹

思路

1. 整個二叉樹用數組存儲
2. 因為先序遍歷它先遍歷根，再遍歷左，左邊沒有跑完是不會去遍歷右邊的，所以遍歷左子樹，就是數組元素每回向后一個，個數-1
3. 遍歷右邊時，就是數組起始位置+左子樹跑到的位置+每次往后走一個，大小就是減去左子樹用掉的個數+每回個數-1
4. 因為要返回遍歷的位置，和遍歷用掉的個數，所以每回都要返回兩個值，用結構體返回。

#include<stdlib.h>
#include<stdio.h>
#include<malloc.h>
#include<string.h>
typedef struct  TreeNode
{struct TreeNode *left;struct TreeNode *right;char val;
}TreeNode;typedef struct Result{TreeNode * root; //構建的樹的根結點int used;      //構建過程中用掉的val個數
} Result;Result CreateTree(char preorder[], int size){if (size == 0){Result result;result.root = NULL;result.used = 0;return result;}char rootVal = preorder[0];if (rootVal == '#'){Result result;result.root = NULL;result.used = 1;    //'#'號在數組中也占一個位置return result;}//根的過程TreeNode * root = (TreeNode *)malloc(sizeof(TreeNode));root->val = rootVal;root->left = root->right = NULL;//左子樹Result left_result = CreateTree(preorder + 1, size - 1);//右子樹Result right_result = CreateTree(preorder + 1 + left_result.used, size - 1 - left_result.used);root->left = left_result.root;root->right = right_result.root;Result result;result.root = root;result.used = 1 + left_result.used + right_result.used;return result;
}void InorderTraversal(TreeNode * root){if (root == NULL){return;}InorderTraversal(root->left);printf("%c", root->val);printf(" ");InorderTraversal(root->right);
}void TestCreateTree()
{char preorder[200];scanf("%s", preorder);int size = strlen(preorder);Result result = CreateTree(preorder, size);InorderTraversal(result.root);
}int main()
{TestCreateTree();return 0;
}

求二叉樹中所有結點的個數

思路

遍歷整棵樹，不是空結點，個數就++

void TreeSize(TreeNode *root, int *size){if (root == NULL){return;}(*size)++;TreeSize(root->left,size);TreeSize(root->right,size);}

思路2

根+左結點個數+右結點個數

int TreeSize2(TreeNode *root){if (root == NULL){return 0;}return 1 + TreeSize2(root->left) + TreeSize2(root->right);
}