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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
/**
* 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
/**
* 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
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
/**
* 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
/**
* 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
递归文章来源：https://www.toymoban.com/news/detail-704371.html
/**
* 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;
}
}
递归文章来源地址https://www.toymoban.com/news/detail-704371.html
/**
* 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;
}
}
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