求二叉树的深度

思路：

分别递归左右子树，深度=左右子树中大的一个+1

/*
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 leftDepth=TreeDepth(pRoot->left);
        int rightDepth=TreeDepth(pRoot->right);
        return leftDepth>rightDepth?leftDepth+1:rightDepth+1;
    }
};

输入一颗二叉树的根节点，判断该树是不是平衡二叉树

思路：

递归每一层都分别算出左右子树的高度，比较，看是否平衡

class Solution {
public:
    bool IsBalanced_Solution(TreeNode* pRoot) {
        //判断平衡，比较高度
        if(pRoot==NULL)
            {
            return true;   //空树是平衡的二叉树
        }
        int leftDep=TreeDepth(pRoot->left);
        int rightDep=TreeDepth(pRoot->right);
        int dif=leftDep-rightDep;
        if(dif>1||dif<-1)
            {
            return false;
        }
        return IsBalanced_Solution(pRoot->left)&&IsBalanced_Solution(pRoot->right);
    }
    
    int TreeDepth(TreeNode* pRoot)
        {
        if(pRoot==NULL)
            {
            return 0;
        }
        int leftDepth=TreeDepth(pRoot->left);
        int rightDepth=TreeDepth(pRoot->right);
        return leftDepth>rightDepth?leftDepth+1:rightDepth+1;
    }
};