平衡二叉树Balanced Binary Tree

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[抄题]:

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.

[思维问题]:

以为要用traverse一直判断,结果发现是没有读题。

[一句话思路]:

定义新的类 来帮助判断。(符合工业规范)

[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

[画图]:

[一刷]:

最后的最大深度是left right中最大的,再+1

[二刷]:

[三刷]:

[四刷]:

[五刷]:

[总结]:

[复杂度]:Time complexity: O(n) Space complexity: O(lgn)

[英文数据结构,为什么不用别的数据结构]:

定义新的类 来帮助判断。(符合工业规范)

[其他解法]:

分治法

[Follow Up]:

[LC给出的题目变变变]:

技术分享图片
/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */


public class Solution {
    /*
     * @param root: The root of binary tree.
     * @return: True if this Binary tree is Balanced, or false.
     */
     class ResultType {
       boolean isBalanced;
       int maxDepth;
       ResultType(boolean isBalanced,int maxDepth) {
           this.isBalanced = isBalanced;
           this.maxDepth = maxDepth;
       }
     };
     
    public boolean isBalanced(TreeNode root) {
        return helper(root).isBalanced;
    }
    
    private ResultType helper (TreeNode root) {
        if (root == null) {
            return new ResultType(true, 0);
        }
        
        ResultType left = helper(root.left);
        ResultType right = helper(root.right);
        
        if (!left.isBalanced || !right.isBalanced) {
            return new ResultType(false, - 1);
        }
        
        else if (Math.abs(left.maxDepth - right.maxDepth) > 1) {
            return new ResultType(false, - 1);
        }
        else {
            return new ResultType(true, Math.max(left.maxDepth,right.maxDepth) + 1);
        }
    }
}
View Code

 

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