标签：for ever problem height 复杂度 bin fine boolean 方法
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.
public class Solution {
public boolean isBalanced(TreeNode root) {
if (root == null) {
return true;
}
boolean left = isBalanced(root.left);
boolean right = isBalanced(root.right);
int l = depth(root.left);
int r = depth(root.right);
return Math.abs(l - r) <= 1 && left && right;
}
public int depth(TreeNode root) {
if (root == null) {
return 0;
}
int left = depth(root.left);
int right = depth(root.right);
return Math.max(left, right) + 1;
}
}
这样的话，每个节点都要求一次depth，depth的时间复杂度是O(n)，所以总共的时间复杂度是O(n^2)。
可以采取自下而上（bottom up）的方法。这样的话，时间复杂度是O(n)。需要注意的是，只要左右有一个是-1，可以直接返回-1。不然的话可能出现两个子树的深度相同，但他们并不是平衡的。
public class Solution {
public boolean isBalanced(TreeNode root) {
return depthDif(root) != -1;
}
public int depthDif(TreeNode root) {
if (root == null) {
return 0;
}
int left = depthDif(root.left);
if (left == -1) {
return -1;
}
int right = depthDif(root.right);
if (right == -1) {
return -1;
}
return Math.abs(left - right) <= 1 ? Math.max(left, right) + 1 : -1;
}
}
标签：for ever problem height 复杂度 bin fine boolean 方法
原文地址：http://www.cnblogs.com/aprilyang/p/6986892.html
