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);    }
    */
}