[LeetCode]110. Balanced Binary Tree

110. 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.

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解法一：递归计算节点的高度

/** * 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:    bool isBalanced(TreeNode* root) {        if(root == NULL)            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 == NULL)            return 0;        return max(height(root->left), height(root->right))+1;    }};

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解法二：后序遍历

/** * 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:    bool isBalanced(TreeNode* root) {        // depth代表当前节点的深度        int depth = 0;        return is(root, depth);    }    // 不仅返回当前节点的左右子树是否平衡的bool值，还要返回当前传入节点的最大深度    bool is(TreeNode* root, int& depth){        // 终止条件,假设空树也是BST        if(root == NULL){            depth = 0;            return true;        }        int left = 0;        int right = 0;        // 如果左右都是平衡的        if(is(root->left, left) && is(root->right, right)){            int dif = abs(left -right);            if(dif <= 1){                depth = max(left, right) + 1;                return true;            }        }        // 不平衡的出现后就不需要管深度了，直接返回就完了，返回一次false，后面全是false        return false;    } };