Q4.1 Implement a function to check if a tree is balanced

Q: Implement a function to check if a tree is balanced. For the purposes of this question, a balanced tree is defined to be a tree such that no two leaf nodes differ in distance from the root by more than one.

A: DFS. 

当前节点为node，计算其左子树的高度left，和右子树的高度right， 如果abs(left-right) > 1,则不是平衡二叉树。

为了节省时间，因为会有子问题重复计算的情况，如果子树不是平衡二叉树，则将其高度置为1

#include <iostream>#include <cmath>using namespace std;struct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};void insert(TreeNode* root, int x){    if(root == NULL){        root = new TreeNode(x);        return;}    if(x < root->val)        insert(root->left, x);    else        insert(root->right, x);}bool isBalance(TreeNode *root) {if (!root) {return true;}return isBalance(root) > -1;}int isBalanced(TreeNode *root) {if (!root) {return 0;}int left = isBalanced(root->left);int right = isBalanced(root->right);if (left < 0 || right < 0 || abs(left-right) > 1) {return -1;}return max(left, right) +１;}int main() {int a[10] = {5,1,2,3,4,6,7,8,9,0};TreeNode *root = NULL;for (int i = 0; i < 10; i++) {insert(root, a[i]);}cout<<isBalance(root)<<endl;}