【leetcode】110.Balanced Binary Tree
2016-11-02 22:14
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题目要求:
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.
即给定一颗二叉树,判断它是不是平衡的
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isBalanced(TreeNode root) {
if(root==null)
{
return true;
}
int l = depth(root.left);
int r = depth(root.right);
if(l-r>1||r-l>1)
{
return false;
}
return isBalanced(root.left)&&isBalanced(root.right);
}
//辅助函数,用来求二叉树的深度
public int depth(TreeNode root)
{
if(root==null)
{
return 0;
}
int left = depth(root.left);
int right = depth(root.right);
return 1+(left>right?left:right);
}
}
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.
即给定一颗二叉树,判断它是不是平衡的
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isBalanced(TreeNode root) {
if(root==null)
{
return true;
}
int l = depth(root.left);
int r = depth(root.right);
if(l-r>1||r-l>1)
{
return false;
}
return isBalanced(root.left)&&isBalanced(root.right);
}
//辅助函数,用来求二叉树的深度
public int depth(TreeNode root)
{
if(root==null)
{
return 0;
}
int left = depth(root.left);
int right = depth(root.right);
return 1+(left>right?left:right);
}
}
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