【劍指Offer學習】【面試題39：二叉樹的深度】
阿新 • • 發佈：2019-01-08
題目一：輸入一棵二叉樹的根結點，求該樹的深度。從根結點到葉子點依次經過的結點（含根、葉結點）形成樹的一條路徑，最長路徑的長度為樹的深度。
二叉樹的結點定義
private static class BinaryTreeNode {
int val;
BinaryTreeNode left;
BinaryTreeNode right;
public BinaryTreeNode() {
}
public BinaryTreeNode(int val) {
this.val = val;
}
}解題思路
如果一棵樹只有一個結點，它的深度為。 如果根結點只有左子樹而沒有右子樹， 那麼樹的深度應該是其左子樹的深度加1，同樣如果根結點只有右子樹而沒有左子樹，那麼樹的深度應該是其右子樹的深度加1. 如果既有右子樹又有左子樹， 那該樹的深度就是其左、右子樹深度的較大值再加1 . 比如在圖6.1 的二叉樹中，根結點為1 的樹有左右兩個子樹，其左右子樹的根結點分別為結點2和3。根結點為2 的左子樹的深度為3 ， 而根結點為3 的右子樹的深度為2，因此根結點為1的樹的深度就是4 。
這個思路用遞迴的方法很容易實現， 只儒對遍歷的程式碼稍作修改即可。
程式碼實現
public static int 題目二：輸入一棵二叉樹的根結點，判斷該樹是不是平衡二叉樹。如果某二叉樹中任意結點的左右子樹的深度相差不超過1 ，那麼它就是一棵平衡二叉樹。
解題思路
解法一：需要重蟹遍歷結點多次的解法
在遍歷樹的每個結點的時候，呼叫函式treeDepth得到它的左右子樹的深度。如果每個結點的左右子樹的深度相差都不超過1 ，按照定義它就是一棵平衡的二叉樹。
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：每個結點只遍歷一次的解法
用後序遍歷的方式遍歷二叉樹的每一個結點，在遍歷到一個結點之前我們就已經遍歷了它的左右子樹。只要在遍歷每個結點的時候記錄它的深度（某一結點的深度等於它到葉節點的路徑的長度），我們就可以一邊遍歷一邊判斷每個結點是不是平衡的。
/**
* 判斷是否是平衡二叉樹，第二種解法
*
* @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;
}完整程式碼
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);
}
/**
* 判斷是否是平衡二叉樹，第一種解法
*
* @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);
}
/**
* 判斷是否是平衡二叉樹，第二種解法
*
* @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();
}
// 完全二叉樹
// 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("----------------");
}
// 不是完全二叉樹，但是平衡二叉樹
// 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("----------------");
}
// 不是平衡二叉樹
// 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("----------------");
}
}