剑指Offer——JZ39.平衡二叉树【DFS】

题目传送门

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题解

• 所求即为：是否有节点的左右子树高度差 > 1
• 树上dfs跑一下即可

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AC-Code

class Solution {
public:
    bool IsBalanced_Solution(TreeNode* pRoot) {
        if(!pRoot)    return true;
        return dfs(pRoot) != -1;
    }
    int dfs(TreeNode* now) {
        if(!now)    return 0;
        int lnum = 0, rnum = 0;	// 别忘了初始化
        if(now->left)
            lnum = dfs(now->left);
        if(now->right)
            rnum = dfs(now->right);
        if(abs(lnum - rnum) > 1 || lnum == -1 || rnum == -1) {
            return -1;
        }
        return max(lnum, rnum) + 1;
    }
};