LeetCode:Balanced Binary Tree
2014-03-28 17:53
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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.
判断是否为平衡二叉树,递归算出某个节点深度,之后用于算深度类似的方法判断是否为平衡二叉树。
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int treeHeight(TreeNode root)
{
if(root==null)return 0;
return Math.max(treeHeight(root.left),treeHeight(root.right))+1;
}
public boolean isBalanced(TreeNode root) {
int height=0;
if(root==null)return true;
return Math.abs(treeHeight(root.left)-treeHeight(root.right))<=1 && isBalanced(root.left) && isBalanced(root.right);
}
}
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.
判断是否为平衡二叉树,递归算出某个节点深度,之后用于算深度类似的方法判断是否为平衡二叉树。
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int treeHeight(TreeNode root)
{
if(root==null)return 0;
return Math.max(treeHeight(root.left),treeHeight(root.right))+1;
}
public boolean isBalanced(TreeNode root) {
int height=0;
if(root==null)return true;
return Math.abs(treeHeight(root.left)-treeHeight(root.right))<=1 && isBalanced(root.left) && isBalanced(root.right);
}
}
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