leetcode-110：判断平衡二叉树 Java

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.
解法1：
1.计算每个节点的高度
2.从根节点开始从上往下遍历，判断每个节点的左右子树是否是平衡的

缺点：每次遍历都要重新计算高度，很多节点的高度都重复计算了，时间复杂度o（n^2）

解法2：

从根节点开始，从上往下遍历，按照中序遍历的思想，从左右子节点向根节点遍历，一依次判断平衡状态，这样根结点可以重复利用已经计算的子节点的高度，只需要依次遍历整棵树。在遇到某个子树非平衡时，能直接结束，返回false。

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
import java.math.*;
public class Solution {public boolean isBalanced(TreeNode root) {if(root == null)return true;int leftH = getHeight(root.left);int rightH = getHeight(root.right);int diff = leftH - rightH;if(diff>1 || diff<-1){return false;}else{return isBalanced(root.left) && isBalanced(root.right);}}int getHeight(TreeNode root){if(root == null){return 0;}return 1+Math.max(getHeight(root.left),getHeight(root.right));}/**public boolean isBalanced(TreeNode root) {if(root == null)return true;if(getHeight(root) == -1){return false;}else{return true;}}public int getHeight(TreeNode root){if(root == null)return 0;int leftH = getHeight(root.left);if(leftH == -1)return -1;int rightH = getHeight(root.right);if(rightH == -1)return -1;if(leftH-rightH > 1 || leftH-rightH < -1)return -1;return 1+(leftH>rightH?leftH:rightH);}*/
}

转载于:https://www.cnblogs.com/wennian/p/5036884.html

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