Tree_Graph 判断是否平衡二叉树 @CareerCup_PHP教程

Implement a function to check if a binary tree is balanced. For the purposes of this question, a balanced tree is defined to be a tree such that the heights of the two subtrees of any node never differ by more than one.

平衡二叉树的定义为：它是一棵空树或它的左右两个子树的高度差的绝对值不超过1， 并且左右两个子树都是一棵平衡二叉树。

思路：

1）先写一个递归的树的高度函数，然后检查子树的高度差是否大于1

2）优化：把检查子树高度差是否大于1的逻辑放在求树的高度的递归函数中，并且遇到非平衡就及时返回。

注：

这道题不同于问一棵树是否平衡（这棵树任意两个叶子结点到根结点的距离之差不大于1）：

喎?http://www.Bkjia.com/kf/ware/vc/" target="_blank" class="keylink">vcD4KPHA+PGJyPgo8L3A+CjxwPsjnyc/NvKOszqrGvbritv6y5sr3o6y1q7K7xr264qGjPC9wPgo8cD7F0LbP0ru/w8r3yse38ca9uuK/ydLUx/PK97XE1+6087jftsi6zdfu0KG437bI1q6y7srHt/G089PaMaGjPC9wPgo8cD7H88r3tcTX7tChuN+2yL/Jss6/vKO6aHR0cDovL2Jsb2cuY3Nkbi5uZXQvZmlnaHRmb3J5b3VyZHJlYW0vYXJ0aWNsZS9kZXRhaWxzLzEyODUxMjMxPC9wPgo8cD48YnI+CjwvcD4KPHA+we3Su9bWveK3qMrHv8nS1NPD1tDQ8rHpwPrH87XDyvfA77XEw7/Su7j20rbX07XEuN+2yKOsyLu687/JtcOhozwvcD4KPHA+ss6/vKO6aHR0cDovL2hhd3N0ZWluLmNvbS9wb3N0cy80LjEuaHRtbDwvcD4KPHA+PGJyPgo8L3A+CjxwPs/Cw+bKx8XQts/Kx7fxxr264rb+subK97XEtPrC66O6PC9wPgo8cD48cHJlIGNsYXNzPQ=="brush:java;">package Tree_Graph; import CtCILibrary.AssortedMethods; import CtCILibrary.TreeNode; public class S4_1 { // 递归判断树是否平衡二叉树 // Time: O(N^2) public static boolean isBalanced(TreeNode root) { if (root == null) { return true; } int heightDiff = getHeight(root.left) - getHeight(root.right); if(Math.abs(heightDiff) > 1) { // 非平衡 return false; } else { return isBalanced(root.left) && isBalanced(root.right); } } // 递归获得树的高度 public static int getHeight(TreeNode root) { if (root == null) { return 0; } return Math.max(getHeight(root.left), getHeight(root.right)) + 1; } // ========================== Improved version 优化版 // 把判断是否平衡的逻辑放在checkHeight函数里，边计算高度， // 边判断是否平衡，如果不平衡，直接返回-1 // Time: O(N), Space: O(H), H: height of tree public static boolean isBalanced2 (TreeNode root) { if (checkHeight(root) == -1) { return false; } else{ return true; } } // 边计算高度边判断是否平衡 public static int checkHeight (TreeNode root) { if (root == null) { return 0; } int leftHeight = checkHeight(root.left); if (leftHeight == -1) { return -1; } int rightHeight = checkHeight(root.right); if (rightHeight == -1) { return -1; } int heightDiff = leftHeight - rightHeight; if (Math.abs(heightDiff) > 1) { return -1; } return Math.max(leftHeight, rightHeight) + 1; } public static void main(String[] args) { // Create balanced tree int[] array = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; TreeNode root = TreeNode.createMinimalBST(array); System.out.println("Root? " + root.data); System.out.println("Is balanced? " + isBalanced(root)); // Could be balanced, actually, but it's very unlikely... TreeNode unbalanced = new TreeNode(10); for (int i = 0; i

http://www.bkjia.com/PHPjc/735868.htmlwww.bkjia.comtruehttp://www.bkjia.com/PHPjc/735868.htmlTechArticleImplement a function to check if a binary tree is balanced. For the purposes of this question, a balanced tree is defined to be a tree such that the heights of the two subtrees of...