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文章目录
- 前言
- 一、力扣104. 二叉树的最大深度
- 二、力扣110. 平衡二叉树
- 三、力扣257. 二叉树的所有路径
- 四、力扣404. 左叶子之和
前言
二叉树中深度指的是根节点到当前节点的节点个数， 二叉树中的高度指的是当前节点到叶子节点的节点个数
可以通前序遍历求深度
通过后序遍历求高度
一、力扣104. 二叉树的最大深度
递归
java">/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
int res;
public int maxDepth(TreeNode root) {
res = 0;
if(root == null)return res;
fun(root, 1);
return res;
}
public void fun(TreeNode root, int cur){
res = res > cur ? res : cur;
if(root.left == null && root.right == null)return;
if(root.left != null){
fun(root.left, cur + 1);
}
if(root.right != null){
fun(root.right, cur + 1);
}
}
}
二、力扣110. 平衡二叉树
递归
java">/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean isBalanced(TreeNode root) {
return fun(root) != -1;
}
public int fun(TreeNode root){
if(root == null)return 0;
int leftHigh = fun(root.left);
if(leftHigh == -1)return -1;
int rightHigh = fun(root.right);
if(rightHigh == -1)return -1;
int res = 0;
if(Math.abs(leftHigh - rightHigh) > 1){
return -1;
}else{
return Math.max(leftHigh, rightHigh) + 1;
}
}
}
迭代
java">class Solution {
/**
* 迭代法，效率较低，计算高度时会重复遍历
* 时间复杂度：O(n^2)
*/
public boolean isBalanced(TreeNode root) {
if (root == null) {
return true;
}
Stack<TreeNode> stack = new Stack<>();
TreeNode pre = null;
while (root!= null || !stack.isEmpty()) {
while (root != null) {
stack.push(root);
root = root.left;
}
TreeNode inNode = stack.peek();
// 右结点为null或已经遍历过
if (inNode.right == null || inNode.right == pre) {
// 比较左右子树的高度差，输出
if (Math.abs(getHeight(inNode.left) - getHeight(inNode.right)) > 1) {
return false;
}
stack.pop();
pre = inNode;
root = null;// 当前结点下，没有要遍历的结点了
} else {
root = inNode.right;// 右结点还没遍历，遍历右结点
}
}
return true;
}
/**
* 层序遍历，求结点的高度
*/
public int getHeight(TreeNode root) {
if (root == null) {
return 0;
}
Deque<TreeNode> deque = new LinkedList<>();
deque.offer(root);
int depth = 0;
while (!deque.isEmpty()) {
int size = deque.size();
depth++;
for (int i = 0; i < size; i++) {
TreeNode poll = deque.poll();
if (poll.left != null) {
deque.offer(poll.left);
}
if (poll.right != null) {
deque.offer(poll.right);
}
}
}
return depth;
}
}
三、力扣257. 二叉树的所有路径
递归
java">/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
List<String> res = new ArrayList<>();
public List<String> binaryTreePaths(TreeNode root) {
List<Integer> li = new ArrayList<>();
if(root == null)return res;
fun(root, li);
return res;
}
public void fun(TreeNode root, List<Integer> li){
li.add(root.val);
if(root.left == null && root.right == null){
StringBuilder sb = new StringBuilder();
for(int i = 0; i < li.size(); i ++){
if(i != li.size()-1){
sb.append(li.get(i)).append("->");
}else{
sb.append(li.get(i));
}
}
res.add(sb.toString());
return;
}
if(root.left != null){
fun(root.left, li);
li.remove(li.size()-1);
}
if(root.right != null){
fun(root.right, li);
li.remove(li.size()-1);
}
}
}
迭代
java">/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<String> binaryTreePaths(TreeNode root) {
Deque<Object> deq = new LinkedList<>();
List<String> res = new ArrayList<>();
if(root == null)return res;
deq.offerLast(root);
deq.offerLast("" + root.val);
while(!deq.isEmpty()){
String path = (String)deq.pollLast();
TreeNode cur = (TreeNode)deq.pollLast();
if(cur.left == null && cur.right == null){
res.add(path);
continue;
}
if(cur.right != null){
deq.offerLast(cur.right);
deq.offerLast(path + "->" + cur.right.val);
}
if(cur.left != null){
deq.offerLast(cur.left);
deq.offerLast(path + "->" + cur.left.val);
}
}
return res;
}
}
四、力扣404. 左叶子之和
递归
java">/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int sumOfLeftLeaves(TreeNode root) {
if(root == null)return 0;
int leftVal = 0, rightVal = 0;
if(root.left != null){
if(root.left.left == null && root.left.right == null){
leftVal = root.left.val;
}
}
leftVal += sumOfLeftLeaves(root.left);
rightVal += sumOfLeftLeaves(root.right);
return leftVal + rightVal;
}
}
递归
java">/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int sumOfLeftLeaves(TreeNode root) {
Deque<TreeNode> deq = new LinkedList<>();
int sum = 0;
if(root == null)return sum;
deq.offerLast(root);
while(!deq.isEmpty()){
int len = deq.size();
for(int i = 0; i < len; i ++){
TreeNode p = deq.pollFirst();
if(p.left != null && p.left.left == null && p.left.right == null){
sum += p.left.val;
}
if(p.left != null)deq.offerLast(p.left);
if(p.right != null)deq.offerLast(p.right);
}
}
return sum;
}
}
