二叉树的深度
题目一:输入一棵二叉树的根结点,求该树的深度。从根结点到叶子点依次经过的结点(含根、叶结点)形成树的一条路径,最长路径的长度为树的深度。
二叉树的结点定义
private static class BinaryTreeNode {
int val;
BinaryTreeNode left;
BinaryTreeNode right;
public BinaryTreeNode() {
}
public BinaryTreeNode(int val) {
this.val = val;
}
}
解题思路
如果一棵树只有一个结点,它的深度为。 如果根结点只有左子树而没有右子树, 那么树的深度应该是其左子树的深度加 1,同样如果根结点只有右子树而没有左子树,那么树的深度应该是其右子树的深度加 1。如果既有右子树又有左子树, 那该树的深度就是其左、右子树深度的较大值再加 1。 比如在图 6.1 的二叉树中,根结点为 1 的树有左右两个子树,其左右子树的根结点分别为结点 2 和 3。根结点为 2 的左子树的深度为 3, 而根结点为 3 的右子树的深度为 2,因此根结点为 1 的树的深度就是 4 。
这个思路用递归的方法很容易实现, 只儒对遍历的代码稍作修改即可。
代码实现
public static int treeDepth(BinaryTreeNode root) {
if (root == null) {
return 0;
}
int left = treeDepth(root.left);
int right = treeDepth(root.right);
return left > right ? (left + 1) : (right + 1);
}
题目二:输入一棵二叉树的根结点,判断该树是不是平衡二叉树。如果某二叉树中任意结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。
解题思路
解法一:需要重蟹遍历结点多次的解法
在遍历树的每个结点的时候,调用函数 treeDepth 得到它的左右子树的深度。如果每个结点的左右子树的深度相差都不超过 1 ,按照定义它就是一棵平衡的二叉树。
public static boolean isBalanced(BinaryTreeNode root) {
if (root == null) {
return true;
}
int left = treeDepth(root.left);
int right = treeDepth(root.right);
int diff = left - right;
if (diff > 1 || diff < -1) {
return false;
}
return isBalanced(root.left) && isBalanced(root.right);
}
解法二:每个结点只遍历一次的解法
用后序遍历的方式遍历二叉树的每一个结点,在遍历到一个结点之前我们就已经遍历了它的左右子树。只要在遍历每个结点的时候记录它的深度(某一结点的深度等于它到叶节点的路径的长度),我们就可以一边遍历一边判断每个结点是不是平衡的。
/**
* 判断是否是平衡二叉树,第二种解法
*
* @param root
* @return
*/
public static boolean isBalanced2(BinaryTreeNode root) {
int[] depth = new int[1];
return isBalancedHelper(root, depth);
}
public static boolean isBalancedHelper(BinaryTreeNode root, int[] depth) {
if (root == null) {
depth[0] = 0;
return true;
}
int[] left = new int[1];
int[] right = new int[1];
if (isBalancedHelper(root.left, left) && isBalancedHelper(root.right, right)) {
int diff = left[0] - right[0];
if (diff >= -1 && diff <= 1) {
depth[0] = 1 + (left[0] > right[0] ? left[0] : right[0]);
return true;
}
}
return false;
}
完整代码
public class Test39 {
private static class BinaryTreeNode {
int val;
BinaryTreeNode left;
BinaryTreeNode right;
public BinaryTreeNode() {
}
public BinaryTreeNode(int val) {
this.val = val;
}
}
public static int treeDepth(BinaryTreeNode root) {
if (root == null) {
return 0;
}
int left = treeDepth(root.left);
int right = treeDepth(root.right);
return left > right ? (left + 1) : (right + 1);
}
/**
* 判断是否是平衡二叉树,第一种解法
*
* @param root
* @return
*/
public static boolean isBalanced(BinaryTreeNode root) {
if (root == null) {
return true;
}
int left = treeDepth(root.left);
int right = treeDepth(root.right);
int diff = left - right;
if (diff > 1 || diff < -1) {
return false;
}
return isBalanced(root.left) && isBalanced(root.right);
}
/**
* 判断是否是平衡二叉树,第二种解法
*
* @param root
* @return
*/
public static boolean isBalanced2(BinaryTreeNode root) {
int[] depth = new int[1];
return isBalancedHelper(root, depth);
}
public static boolean isBalancedHelper(BinaryTreeNode root, int[] depth) {
if (root == null) {
depth[0] = 0;
return true;
}
int[] left = new int[1];
int[] right = new int[1];
if (isBalancedHelper(root.left, left) && isBalancedHelper(root.right, right)) {
int diff = left[0] - right[0];
if (diff >= -1 && diff <= 1) {
depth[0] = 1 + (left[0] > right[0] ? left[0] : right[0]);
return true;
}
}
return false;
}
public static void main(String[] args) {
test1();
test2();
test3();
test4();
}
// 完全二叉树
// 1
// / \
// 2 3
// /\ / \
// 4 5 6 7
private static void test1() {
BinaryTreeNode n1 = new BinaryTreeNode(1);
BinaryTreeNode n2 = new BinaryTreeNode(1);
BinaryTreeNode n3 = new BinaryTreeNode(1);
BinaryTreeNode n4 = new BinaryTreeNode(1);
BinaryTreeNode n5 = new BinaryTreeNode(1);
BinaryTreeNode n6 = new BinaryTreeNode(1);
BinaryTreeNode n7 = new BinaryTreeNode(1);
n1.left = n2;
n1.right = n3;
n2.left = n4;
n2.right = n5;
n3.left = n6;
n3.right = n7;
System.out.println(isBalanced(n1));
System.out.println(isBalanced2(n1));
System.out.println("----------------");
}
// 不是完全二叉树,但是平衡二叉树
// 1
// / \
// 2 3
// /\ \
// 4 5 6
// /
// 7
private static void test2() {
BinaryTreeNode n1 = new BinaryTreeNode(1);
BinaryTreeNode n2 = new BinaryTreeNode(1);
BinaryTreeNode n3 = new BinaryTreeNode(1);
BinaryTreeNode n4 = new BinaryTreeNode(1);
BinaryTreeNode n5 = new BinaryTreeNode(1);
BinaryTreeNode n6 = new BinaryTreeNode(1);
BinaryTreeNode n7 = new BinaryTreeNode(1);
n1.left = n2;
n1.right = n3;
n2.left = n4;
n2.right = n5;
n5.left = n7;
n3.right = n6;
System.out.println(isBalanced(n1));
System.out.println(isBalanced2(n1));
System.out.println("----------------");
}
// 不是平衡二叉树
// 1
// / \
// 2 3
// /\
// 4 5
// /
// 7
private static void test3() {
BinaryTreeNode n1 = new BinaryTreeNode(1);
BinaryTreeNode n2 = new BinaryTreeNode(1);
BinaryTreeNode n3 = new BinaryTreeNode(1);
BinaryTreeNode n4 = new BinaryTreeNode(1);
BinaryTreeNode n5 = new BinaryTreeNode(1);
BinaryTreeNode n6 = new BinaryTreeNode(1);
BinaryTreeNode n7 = new BinaryTreeNode(1);
n1.left = n2;
n1.right = n3;
n2.left = n4;
n2.right = n5;
n5.left = n7;
System.out.println(isBalanced(n1));
System.out.println(isBalanced2(n1));
System.out.println("----------------");
}
// 1
// /
// 2
// /
// 3
// /
// 4
// /
// 5
private static void test4() {
BinaryTreeNode n1 = new BinaryTreeNode(1);
BinaryTreeNode n2 = new BinaryTreeNode(1);
BinaryTreeNode n3 = new BinaryTreeNode(1);
BinaryTreeNode n4 = new BinaryTreeNode(1);
BinaryTreeNode n5 = new BinaryTreeNode(1);
BinaryTreeNode n6 = new BinaryTreeNode(1);
BinaryTreeNode n7 = new BinaryTreeNode(1);
n1.left = n2;
n2.left = n3;
n3.left = n4;
n4.left = n5;
System.out.println(isBalanced(n1));
System.out.println(isBalanced2(n1));
System.out.println("----------------");
}
// 1
// \
// 2
// \
// 3
// \
// 4
// \
// 5
private static void test5() {
BinaryTreeNode n1 = new BinaryTreeNode(1);
BinaryTreeNode n2 = new BinaryTreeNode(1);
BinaryTreeNode n3 = new BinaryTreeNode(1);
BinaryTreeNode n4 = new BinaryTreeNode(1);
BinaryTreeNode n5 = new BinaryTreeNode(1);
BinaryTreeNode n6 = new BinaryTreeNode(1);
BinaryTreeNode n7 = new BinaryTreeNode(1);
n1.right = n2;
n2.right = n3;
n3.right = n4;
n4.right = n5;
System.out.println(isBalanced(n1));
System.out.println(isBalanced2(n1));
System.out.println("----------------");
}
}
运行结果
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