LeetCode:Balanced Binary Tree(平衡二叉树的判断)

LeetCode:Balanced Binary Tree(平衡二叉树的判断)

1、题目:
Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

2、代码:

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */
class Solution {
public:
    //Depth-First Search
    bool inorder(TreeNode* root,int &height)
    {
       if(root)
       {
           int hleft=0,hright=0;
           if(inorder(root->left,hleft)&&inorder(root->right,hright)&&abs(hright-hleft)<=1)
           {
               height=max(hleft,hright)+1;
           }
           else
           {
               return false;
           }
       }
       return true;
    }

    bool isBalanced(TreeNode* root) {
        if(!root)   return true;
        int height=0;
        return inorder(root,height);
    }
};

3、总结:
从昨天写到今天,总算有点突破,AC了。
A、定义:它是一 棵空树或它的左右两个子树的高度差的绝对值不超过1,并且左右两个子树都是一棵平衡二叉树。跟题目里面描述的类似。
B、既然是高度相关,那首当其冲应该深度优先搜索。按照遍历类似,应该选择后序遍历。
C、得到左右子树高度,并保证两个子树都是平衡的,算出当前结点的高度max(hleft,hright)+1,别忘记+1!!!

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