【剑指offer】面试题55（2）：平衡二叉树

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题目

输入一棵二叉树的根节点，判断该树是不是平衡二叉树。
如果某二叉树中的任意节点的左、右子树的深度相差不超过1，那么它就是一棵平衡二叉树。

思路

我们用后序遍历的方式遍历二叉树的每个节点，那么在遍历到一个节点之前我们就已经遍历了它的左、右子树。
只要在遍历每个节点的时候记录它的深度，我们就可以一边遍历一边判断每个节点是不是平衡的。

ps：根节点为null时是平衡二叉树

代码

/** * 题目： * 输入一棵二叉树的根节点，判断该树是不是平衡二叉树。 * 如果某二叉树中的任意节点的左、右子树的深度相差不超过1，那么它就是一棵平衡二叉树。 *  * ps：根节点为null时返回true *  * @author peige */public class _55_02_BalancedBinaryTree {    public static class TreeNode {        public int val = 0;        public TreeNode left = null;        public TreeNode right = null;        public TreeNode(int val) {            this.val = val;        }    }    public boolean IsBalanced_Solution(TreeNode root) {        if(root == null)            return true;        int res = treeDepth(root);        if(res == -1)            return false;        return true;    }    /**     * 一旦出现不平衡的节点，开始递归返回-1      */    public int treeDepth(TreeNode root) {        if(root == null)            return 0;        int leftDepth = treeDepth(root.left);        if(leftDepth == -1)            return -1;        int rightDepth = treeDepth(root.right);        if(rightDepth == -1)            return -1;        // 该节点是否平衡        if(leftDepth - rightDepth > 1 || leftDepth - rightDepth < -1)            return -1;        // 加上根节点的长度（后序遍历）        return leftDepth > rightDepth ? leftDepth + 1: rightDepth + 1;    }}

测试

public class _55_02_Test {    public static void main(String[] args) {        test1();        test2();        test3();        test4();    }    /**     * 功能测试     * 树1：     *         1     *       /   \     *      2     3     *     / \   / \     *    4   5 6   7     *   /     *  8      *       * 树2：     *         1     *       /   \     *      2     3     *     / \   / \     *    4   5 6   7     *   /   / \     *  8   9  10     *       \     *       11     *       */    private static void test1() {        TreeNode node1 = new TreeNode(1);        TreeNode node2 = new TreeNode(2);        TreeNode node3 = new TreeNode(3);        TreeNode node4 = new TreeNode(4);        TreeNode node5 = new TreeNode(5);        TreeNode node6 = new TreeNode(6);        TreeNode node7 = new TreeNode(7);        TreeNode node8 = new TreeNode(8);        node1.left = node2;        node1.right = node3;        node2.left = node4;        node2.right = node5;        node3.left = node6;        node3.right = node7;        node4.left = node8;        boolean b = new _55_02_BalancedBinaryTree().IsBalanced_Solution(node1);        MyTest.equal(b, true);        TreeNode node9 = new TreeNode(9);        TreeNode node10 = new TreeNode(10);        TreeNode node11 = new TreeNode(11);        node5.left = node9;        node5.right = node10;        node9.right = node11;        b = new _55_02_BalancedBinaryTree().IsBalanced_Solution(node1);        MyTest.equal(b, false);    }    private static void test2() {        TreeNode node1 = new TreeNode(1);        _55_02_BalancedBinaryTree bbt = new _55_02_BalancedBinaryTree();        boolean b = bbt.IsBalanced_Solution(node1);        MyTest.equal(b, true);    }    private static void test3() {        _55_02_BalancedBinaryTree bbt = new _55_02_BalancedBinaryTree();        boolean b = bbt.IsBalanced_Solution(null);        MyTest.equal(b, false);    }    /**     * 提交的时候没过的case：     *             1     *            /     *           2       *          /     *         3     *        /     *       4     *      /     *     5          */    private static void test4() {        TreeNode node1 = new TreeNode(1);        TreeNode node2 = new TreeNode(2);        TreeNode node3 = new TreeNode(3);        TreeNode node4 = new TreeNode(4);        TreeNode node5 = new TreeNode(5);        node1.left = node2;        node2.left = node3;        node3.left = node4;        node4.left = node5;        _55_02_BalancedBinaryTree bbt = new _55_02_BalancedBinaryTree();        boolean b =bbt.IsBalanced_Solution(node1);        MyTest.equal(b, false);    }}

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