LeetCode 110. Balanced Binary Tree(平衡树)

题意：判断二叉树是否为平衡树。左右子树高度差最多为1的树是平衡树。

法一：

（1）先判断左右子树是否为平衡树

（2）再判断左右子树是否高度差最多为1

（3）更新树高(root->val记录当前树高)

/**
 * 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(!isBalanced(root -> left)) return false;
        if(!isBalanced(root -> right)) return false;
        int leftval, rightval;
        if(root -> left == NULL) leftval = 0;
        else leftval = root -> left -> val;
        if(root -> right == NULL) rightval = 0;
        else rightval = root -> right -> val;
        if(abs(leftval - rightval) > 1) return false;
        root -> val = max(leftval, rightval) + 1;
        return true;
    }
};

法二：借助求树高判断是否为平衡树。

/**
 * 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 ans = true;
    int maxDepth(TreeNode* root){
        if(root == NULL) return 0;
        int l = maxDepth(root -> left);
        int r = maxDepth(root -> right);
        if(abs(l - r) > 1) ans = false;
        return max(l, r) + 1;
    }
    bool isBalanced(TreeNode* root) {
        maxDepth(root);
        return ans;
    }
};