110 - Balanced Binary Tree

Given a binary tree, determine if it is height-balanced

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: true

Example 2:

Input: root = [1,2,2,3,3,null,null,4,4]
Output: false

思路

后序: 左右中 (向上遞歸)
前序: 中左右 (向下遞歸)

Solution

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int getHeight( TreeNode* root ) {
        if ( root == NULL ) return 0;
        int leftHeight = getHeight( root -> left ); // 左
        if ( leftHeight == -1 ) return -1;
        int rightHeight = getHeight( root -> right ); // 右
        if ( rightHeight == -1 ) return -1;
        int result;
        if ( abs( leftHeight - rightHeight ) > 1 ) result = -1; // 中
        else result = max( leftHeight, rightHeight ) + 1; // 加上本身節點高度(1)
        return result;
    }

    bool isBalanced(TreeNode* root) {
        int result = getHeight(root);
        if ( result == -1 ) return false;
        return true;