[LeetCode] 110. Balanced Binary Tree

Balanced Binary Tree

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.

解析

判断一颗树是不是平衡树，获得左右子树的高度，绝对值不大于1就好。

解法1：递归

直接每个节点递归判断左右子树高度差绝对值。

class Solution {
public:
    bool isBalanced(TreeNode* root) {
        if(!root) return true;
        if(abs(height(root->left) - height(root->right)) > 1) return false;
        return isBalanced(root->left) && isBalanced(root->right);
    }

    int height(TreeNode* root){
        if(!root) return 0;
        return max(height(root->left), height(root->right)) + 1;
    }
};

解法2：checkdepth判断子树是否平衡

如果子树不平衡，则不计算具体的深度，而是直接返回-1。优化后为：对于每一个节点，我们通过checkDepth方法递归获得左右子树的深度，如果子树是平衡的，则返回真实的深度，若不平衡，直接返回-1，此方法时间复杂度O(N)，空间复杂度O(H)。

class Solution {
public:
    bool isBalanced(TreeNode* root) {
        if(checkdepth(root) == -1) return false;
        else return true;
    }

    int checkdepth(TreeNode* root){
        if(!root) return 0;
        int left = checkdepth(root->left);
        if(left == -1) return -1;
        int right = checkdepth(root->right);
        if(right == -1) return -1;
        if(abs(left - right) > 1) return -1;
        else return 1 + max(left, right);
    }
};

参考

https://www.cnblogs.com/grandyang/p/4045660.html