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

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本文介绍了一种使用深度优先搜索(DFS)的方法来判断一棵二叉树是否为平衡树。平衡树定义为任意两个叶子节点到根的距离之差不超过1的树。通过递归计算左右子树的高度并比较，若相差超过1则该树不平衡。

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;
}