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.
给定一棵平衡二叉树,判断它是否是高度平衡的。一棵高度平衡的二叉树是左右子树的高度相差不超过1,对其左右子树也是如此。
递归分治法求解。
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
算法实现类
public class Solution {
private int minDepth = 0;
private int maxDepth = 0;
public boolean isBalanced(TreeNode root) {
if (root == null) {
return true;
}
int left = depth(root.left);
int right = depth(root.right);
if (left - right > 1 || left - right < -1) {
return false;
} else {
return isBalanced(root.left) && isBalanced(root.right);
}
}
/**
* 求树的高度
* @param n 树的根结点
* @return 树的高度
*/
private int depth(TreeNode n) {
if (n == null) {
return 0;
} if (n.left == null && n.right == null) {
return 1;
} else {
int left = depth(n.left);
int right = depth(n.right);
return 1 + (left > right ? left : right);
}
}
}
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【LeetCode-面试算法经典-Java实现】【110-Balanced Binary Tree(平衡二叉树)】
原文地址:http://blog.csdn.net/derrantcm/article/details/47414041