标签：不用 ble help color private math tty eterm public
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.
Example
Given binary tree A = {3,9,20,#,#,15,7}, B = {3,#,20,15,7}
A) 3 B) 3
/ \ 9 20 20
/ \ / 15 7 15 7
The binary tree A is a height-balanced binary tree, but B is not.
1. 分治+遍历，用全局变量。注意一下根据定义，空树null也是一颗平衡二叉树。
/** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /* * @param root: The root of binary tree. * @return: True if this Binary tree is Balanced, or false. */ private boolean isBalanceTree; public boolean isBalanced(TreeNode root) { // write your code here isBalanceTree = true; height(root); return isBalanceTree; } private int height(TreeNode root) { if (root == null) { return 0; } int leftHeight = height(root.left); int rightHeight = height(root.right); if (Math.abs(leftHeight - rightHeight) > 1) { isBalanceTree = false; } return Math.max(leftHeight, rightHeight) + 1; } }
2.分治，不用全局变量用ResultType。
/** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /* * @param root: The root of binary tree. * @return: True if this Binary tree is Balanced, or false. */ private class ResultType{ public int height; public boolean isBalanceTree; public ResultType(int height, boolean isBalanceTree) { this.height = height; this.isBalanceTree = isBalanceTree; } } public boolean isBalanced(TreeNode root) { // write your code here ResultType result = helper(root); return result.isBalanceTree; } private ResultType helper(TreeNode root) { ResultType result = new ResultType(0, true); if (root == null) { return result; } ResultType left = helper(root.left); ResultType right = helper(root.right); if (!left.isBalanceTree || !right.isBalanceTree || Math.abs(left.height - right.height) > 1) { result.isBalanceTree = false; } result.height = Math.max(left.height, right.height) + 1; return result; } }
lintcode93- Balanced Binary Tree- easy
标签：不用 ble help color private math tty eterm public
原文地址：http://www.cnblogs.com/jasminemzy/p/7639475.html
