平衡二叉树

平衡二叉树

题目描述

输入一棵二叉树，判断该二叉树是否是平衡二叉树。

不是很懂, 一边递归一边判断, 可以消除节点遍历两次

class Solution {
public:
    bool IsBalanced_Solution(TreeNode *pRoot, int *pDepth) {
        if (nullptr == pRoot) {
            *pDepth = 0;
            return true;
        }
        int left = 0;
        int right = 0;
        if ((IsBalanced_Solution(pRoot->left, &left)) 
            && IsBalanced_Solution(pRoot->right, &right)) {
            int dif = left - right;
            if ((dif >= -1) && (dif <= 1)) {
                *pDepth = 1 + (left > right ? left : right);
                return true;
            }
        }
        
        return false;
    }
    
    bool IsBalanced_Solution(TreeNode* pRoot) {
        int depth = 0;
        
        return IsBalanced_Solution(pRoot, &depth);
    }
};

递归判断布尔变量方法学学, 好像用到两次了, 节点遍历两次

class Solution {
public:
    int TreeDepth(TreeNode* pRoot)
    {
        int rightHigh = 1;
        int leftHigh = 1;
        
        if (nullptr == pRoot) {
            return 0;
        }
        
        if (nullptr != pRoot->left) {
            rightHigh += TreeDepth(pRoot->left);
        }
        
        if (nullptr != pRoot->right) {
            leftHigh += TreeDepth(pRoot->right);
        }
        
        return rightHigh > leftHigh ? rightHigh : leftHigh;
    }
    
    
    bool IsBalanced_Solution(TreeNode* pRoot) {
        if (nullptr == pRoot) {
            return true;
        }
        int left = TreeDepth(pRoot->left);
        int right = TreeDepth(pRoot->right);
        int dif = left - right;
        
        if (dif < -1 || dif >1) {
            return false;
        }
        return IsBalanced_Solution(pRoot->left) && IsBalanced_Solution(pRoot->right);
    }
};