第六章 二叉树 part05
今日内容
● 513.找树左下角的值
● 112. 路径总和 113.路径总和ii
● 106.从中序与后序遍历序列构造二叉树 105.从前序与中序遍历序列构造二叉树
详细布置
找树左下角的值
本地递归偏难，反而迭代简单属于模板题， 两种方法掌握一下
题目链接/文章讲解/视频讲解：https://programmercarl.com/0513.%E6%89%BE%E6%A0%91%E5%B7%A6%E4%B8%8B%E8%A7%92%E7%9A%84%E5%80%BC.html
路径总和
本题 又一次设计要回溯的过程，而且回溯的过程隐藏的还挺深，建议先看视频来理解
112. 路径总和，和 113. 路径总和ii 一起做了。 优先掌握递归法。
题目链接/文章讲解/视频讲解：https://programmercarl.com/0112.%E8%B7%AF%E5%BE%84%E6%80%BB%E5%92%8C.html
从中序与后序遍历序列构造二叉树
本题算是比较难的二叉树题目了，大家先看视频来理解。
106.从中序与后序遍历序列构造二叉树，105.从前序与中序遍历序列构造二叉树 一起做，思路一样的
题目链接/文章讲解/视频讲解：https://programmercarl.com/0106.%E4%BB%8E%E4%B8%AD%E5%BA%8F%E4%B8%8E%E5%90%8E%E5%BA%8F%E9%81%8D%E5%8E%86%E5%BA%8F%E5%88%97%E6%9E%84%E9%80%A0%E4%BA%8C%E5%8F%89%E6%A0%91.html
往日任务
● day 1 任务以及具体安排：https://docs.qq.com/doc/DUG9UR2ZUc3BjRUdY
● day 2 任务以及具体安排：https://docs.qq.com/doc/DUGRwWXNOVEpyaVpG
● day 3 任务以及具体安排：https://docs.qq.com/doc/DUGdqYWNYeGhlaVR6
● day 4 任务以及具体安排：https://docs.qq.com/doc/DUFNjYUxYRHRVWklp
● day 5 周日休息
● day 6 任务以及具体安排：https://docs.qq.com/doc/DUEtFSGdreWRuR2p4
● day 7 任务以及具体安排：https://docs.qq.com/doc/DUElCb1NyTVpXa0Jj
● day 8 任务以及具体安排：https://docs.qq.com/doc/DUGdsY2JFaFhDRVZH
● day 9 任务以及具体安排：https://docs.qq.com/doc/DUHVXSnZNaXpVUHN4
● day 10 任务以及具体安排：https://docs.qq.com/doc/DUElqeHh3cndDbW1Q
●day 11 任务以及具体安排：https://docs.qq.com/doc/DUHh6UE5hUUZOZUd0
●day 12 周日休息
●day 13 任务以及具体安排：https://docs.qq.com/doc/DUHNpa3F4b2dMUWJ3
●day 14 任务以及具体安排：https://docs.qq.com/doc/DUHRtdXZZSWFkeGdE
●day 15 任务以及具体安排：https://docs.qq.com/doc/DUHN0ZVJuRmVYeWNv
●day 16 任务以及具体安排：https://docs.qq.com/doc/DUHBQRm1aSWR4T2NK
●day 17 任务以及具体安排：https://docs.qq.com/doc/DUFpXY3hBZkpabWFY目录
0513_找树左下角的值
package com.question.solve.leetcode.programmerCarl2._07_binaryTrees;
import java.util.Deque;
import java.util.LinkedList;
public class _0513_找树左下角的值 {
}
/**
* 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 Solution0513 {
public int findBottomLeftValue(TreeNode root) {//迭代法
int res = 0;
if (root == null) {
return res;
}
Deque<TreeNode> deuqe = new LinkedList<>();
deuqe.offer(root);
while (!deuqe.isEmpty()) {
int size = deuqe.size();
for (int i = 0; i < size; i++) {
TreeNode poll = deuqe.poll();
if (i == 0) {
res = poll.val;
}
if (poll.left != null) {
deuqe.offer(poll.left);
}
if (poll.right != null) {
deuqe.offer(poll.right);
}
}
}
return res;
}
}
class Solution0513_2 {//递归法
private int Deep = -1;
private int value = 0;
public int findBottomLeftValue(TreeNode root) {
value = root.val;
findLeftValue(root, 0);
return value;
}
private void findLeftValue(TreeNode root, int deep) {
if (root == null) return;
if (root.left == null && root.right == null) {
if (deep > Deep) {
value = root.val;
Deep = deep;
}
}
if (root.left != null) findLeftValue(root.left, deep + 1);
if (root.right != null) findLeftValue(root.right, deep + 1);
}
}0112_路径总和
package com.question.solve.leetcode.programmerCarl2._07_binaryTrees;
import java.util.ArrayList;
import java.util.Stack;
public class _0112_路径总和 {
}
/**
* 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 Solution0112 {
public boolean hasPathSum(TreeNode root, int targetSum) {
ArrayList<Integer> paths = new ArrayList<>();
Boolean flag = false; //使用 Boolean包装类 来存储flag的状态
traversal(root, paths, targetSum, flag);//flag不能是基本数据类型，
return flag; //简化返回逻辑，直接返回flag
// int[] flag = new int[1]; //使用长度为1的数组来存储flag的状态
// //Integer flag = 0; //使用 Integer包装类 来存储flag的状态
// //Boolean flag = false; //使用 Boolean包装类 来存储flag的状态
// traversal(root, paths, targetSum, flag);
// return flag[0] == 1; //简化返回逻辑
}
private void traversal(TreeNode root, ArrayList<Integer> paths, int targetSum, Boolean flag) {
if (root == null || flag) { //如果 flag 已经为 true，则不再进行遍历
return;
}
paths.add(root.val);
if (root.left == null && root.right == null) {
int sum = 0;
for (int x : paths) {
sum += x;
}
if (sum == targetSum) {
flag = true;
return;
}
}
if (root.left != null) {
traversal(root.left, paths, targetSum, flag);
paths.remove(paths.size() - 1);
}
if (root.right != null) {
traversal(root.right, paths, targetSum, flag);
paths.remove(paths.size() - 1);
}
}
}
class Solution0112_2 {
public boolean haspathsum(TreeNode root, int targetsum) {
if (root == null) {
return false;
}
targetsum -= root.val;
//叶子结点
if (root.left == null && root.right == null) {
return targetsum == 0;
}
if (root.left != null) {
boolean left = haspathsum(root.left, targetsum);
if (left) { //已经找到
return true;
}
}
if (root.right != null) {
boolean right = haspathsum(root.right, targetsum);
if (right) { //已经找到
return true;
}
}
return false;
}
public boolean haspathsum2(TreeNode root, int targetsum) {//lc112 简洁方法
if (root == null) return false; //为空退出
//叶子节点判断是否符合
if (root.left == null && root.right == null) return root.val == targetsum;
//求两侧分支的路径和
return haspathsum2(root.left, targetsum - root.val) || haspathsum(root.right, targetsum - root.val);
}
}
class Solution0112_3 {
public boolean hasPathSum(TreeNode root, int targetSum) {
if (root == null) return false;
Stack<TreeNode> stack1 = new Stack<>();
Stack<Integer> stack2 = new Stack<>();
stack1.push(root);
stack2.push(root.val);
while (!stack1.isEmpty()) {
int size = stack1.size();
for (int i = 0; i < size; i++) {
TreeNode node = stack1.pop();
int sum = stack2.pop();
//如果该节点是叶子节点了，同时该节点的路径数值等于sum，那么就返回true
if (node.left == null && node.right == null && sum == targetSum) {
return true;
}
//右节点，压进去一个节点的时候，将该节点的路径数值也记录下来
if (node.right != null) {
stack1.push(node.right);
stack2.push(sum + node.right.val);
}
//左节点，压进去一个节点的时候，将该节点的路径数值也记录下来
if (node.left != null) {
stack1.push(node.left);
stack2.push(sum + node.left.val);
}
}
}
return false;
}
public boolean hasPathSum2(TreeNode root, int targetSum) {
Stack<TreeNode> treeNodeStack = new Stack<>();
Stack<Integer> sumStack = new Stack<>();
if (root == null)
return false;
treeNodeStack.add(root);
sumStack.add(root.val);
while (!treeNodeStack.isEmpty()) {
TreeNode curr = treeNodeStack.peek();
int tempsum = sumStack.pop();
if (curr != null) {
treeNodeStack.pop();
treeNodeStack.add(curr);
treeNodeStack.add(null);
sumStack.add(tempsum);
if (curr.right != null) {
treeNodeStack.add(curr.right);
sumStack.add(tempsum + curr.right.val);
}
if (curr.left != null) {
treeNodeStack.add(curr.left);
sumStack.add(tempsum + curr.left.val);
}
} else {
treeNodeStack.pop();
TreeNode temp = treeNodeStack.pop();
if (temp.left == null && temp.right == null && tempsum == targetSum)
return true;
}
}
return false;
}
}0113_路径总和ii
package com.question.solve.leetcode.programmerCarl2._07_binaryTrees;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Stack;
public class _0113_路径总和II {
}
/**
* 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 Solution0113 {
public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) {//非空判断
return res;
}
List<Integer> paths = new ArrayList<>();
traversal(root, targetSum, res, paths);
return res;
}
private void traversal(TreeNode root, int targetSum, List<List<Integer>> res, List<Integer> paths) {
if (root == null) {
return;
}
paths.add(root.val);
if (root.left == null && root.right == null) {
int sum = paths.stream().mapToInt(Integer::intValue).sum();
if (targetSum == sum) {
res.add(new ArrayList<>(paths));
}
}
if (root.left != null) {
traversal(root.left, targetSum, res, paths);
paths.remove(paths.size() - 1);
}
if (root.right != null) {
traversal(root.right, targetSum, res, paths);
paths.remove(paths.size() - 1);
}
}
}
class Solution0113_2 {
public List<List<Integer>> pathsum(TreeNode root, int targetsum) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) return res;//非空判断
List<Integer> path = new LinkedList<>();
preOrderDfs(root, targetsum, res, path);
return res;
}
public void preOrderDfs(TreeNode root, int targetsum, List<List<Integer>> res, List<Integer> path) {
path.add(root.val);
if (root.left == null && root.right == null) {//遇到了叶子节点
if (targetsum - root.val == 0) {//找到了和为targetSum的路径
res.add(new ArrayList<>(path));
}
return;//如果和不为targetSum，返回
}
if (root.left != null) {
preOrderDfs(root.left, targetsum - root.val, res, path);
path.remove(path.size() - 1); //回溯
}
if (root.right != null) {
preOrderDfs(root.right, targetsum - root.val, res, path);
path.remove(path.size() - 1); //回溯
}
}
}
class Solution0113_3 {
List<List<Integer>> result;
LinkedList<Integer> path;
public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
result = new LinkedList<>();
path = new LinkedList<>();
travesal(root, targetSum);
return result;
}
private void travesal(TreeNode root, int count) {
if (root == null) return;
path.offer(root.val);
count -= root.val;
if (root.left == null && root.right == null && count == 0) {
result.add(new LinkedList<>(path));
}
travesal(root.left, count);
travesal(root.right, count);
path.removeLast();//回溯
}
}
class Solution0113_4 {
public List<List<Integer>> pathSum(TreeNode root, int targetSum) {//DFS统一迭代法
List<List<Integer>> result = new ArrayList<>();
Stack<TreeNode> nodeStack = new Stack<>();
Stack<Integer> sumStack = new Stack<>();
Stack<ArrayList<Integer>> pathStack = new Stack<>();
if (root == null)
return result;
nodeStack.add(root);
sumStack.add(root.val);
pathStack.add(new ArrayList<>());
while (!nodeStack.isEmpty()) {
TreeNode currNode = nodeStack.peek();
int currSum = sumStack.pop();
ArrayList<Integer> currPath = pathStack.pop();
if (currNode != null) {
nodeStack.pop();
nodeStack.add(currNode);
nodeStack.add(null);
sumStack.add(currSum);
currPath.add(currNode.val);
pathStack.add(new ArrayList(currPath));
if (currNode.right != null) {
nodeStack.add(currNode.right);
sumStack.add(currSum + currNode.right.val);
pathStack.add(new ArrayList(currPath));
}
if (currNode.left != null) {
nodeStack.add(currNode.left);
sumStack.add(currSum + currNode.left.val);
pathStack.add(new ArrayList(currPath));
}
} else {
nodeStack.pop();
TreeNode temp = nodeStack.pop();
if (temp.left == null && temp.right == null && currSum == targetSum)
result.add(new ArrayList(currPath));
}
}
return result;
}
}0106_从中序与后序遍历序列构造二叉树
package com.question.solve.leetcode.programmerCarl2._07_binaryTrees;
import java.util.Deque;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Map;
public class _0106_从中序与后序遍历序列构造二叉树 {
}
/**
* 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 Solution0106 {
Map<Integer, Integer> map; //方便根据数值查找位置
public TreeNode buildTree(int[] inorder, int[] postorder) {
map = new HashMap<>();
for (int i = 0; i < inorder.length; i++) { //用map保存中序序列的数值对应位置
map.put(inorder[i], i);
}
return findNode(inorder, 0, inorder.length, postorder, 0, postorder.length); // 前闭后开
}
public TreeNode findNode(int[] inorder, int inBegin, int inEnd, int[] postorder, int postBegin, int postEnd) {
//参数里的范围都是前闭后开
if (inBegin >= inEnd || postBegin >= postEnd) { // 不满足左闭右开，说明没有元素，返回空树
return null;
}
int rootIndex = map.get(postorder[postEnd - 1]); // 找到后序遍历的最后一个元素在中序遍历中的位置
TreeNode root = new TreeNode(inorder[rootIndex]); // 构造结点
int lenOfLeft = rootIndex - inBegin; // 保存中序左子树个数，用来确定后序数列的个数
root.left = findNode(inorder, inBegin, rootIndex,
postorder, postBegin, postBegin + lenOfLeft);
root.right = findNode(inorder, rootIndex + 1, inEnd,
postorder, postBegin + lenOfLeft, postEnd - 1);
return root;
}
}
class Solution0106_2 {
public TreeNode buildTree(int[] inorder, int[] postorder) {
if (postorder.length == 0 || inorder.length == 0)
return null;
return buildHelper(inorder, 0, inorder.length, postorder, 0, postorder.length);
}
private TreeNode buildHelper(int[] inorder, int inorderStart, int inorderEnd, int[] postorder, int postorderStart, int postorderEnd) {
if (postorderStart == postorderEnd)
return null;
int rootVal = postorder[postorderEnd - 1];
TreeNode root = new TreeNode(rootVal);
int middleIndex;
for (middleIndex = inorderStart; middleIndex < inorderEnd; middleIndex++) {
if (inorder[middleIndex] == rootVal)
break;
}
int leftInorderStart = inorderStart;
int leftInorderEnd = middleIndex;
int rightInorderStart = middleIndex + 1;
int rightInorderEnd = inorderEnd;
int leftPostorderStart = postorderStart;
int leftPostorderEnd = postorderStart + (middleIndex - inorderStart);
int rightPostorderStart = leftPostorderEnd;
int rightPostorderEnd = postorderEnd - 1;
root.left = buildHelper(inorder, leftInorderStart, leftInorderEnd, postorder, leftPostorderStart, leftPostorderEnd);
root.right = buildHelper(inorder, rightInorderStart, rightInorderEnd, postorder, rightPostorderStart, rightPostorderEnd);
return root;
}
}
class Solution0106_3 {
public TreeNode buildTree(int[] inorder, int[] postorder) {
if (postorder == null || postorder.length == 0) {
return null;
}
TreeNode root = new TreeNode(postorder[postorder.length - 1]);
Deque<TreeNode> stack = new LinkedList<TreeNode>();
stack.push(root);
int inorderIndex = inorder.length - 1;
for (int i = postorder.length - 2; i >= 0; i--) {
int postorderVal = postorder[i];
TreeNode node = stack.peek();
if (node.val != inorder[inorderIndex]) {
node.right = new TreeNode(postorderVal);
stack.push(node.right);
} else {
while (!stack.isEmpty() && stack.peek().val == inorder[inorderIndex]) {
node = stack.pop();
inorderIndex--;
}
node.left = new TreeNode(postorderVal);
stack.push(node.left);
}
}
return root;
}
}0105_从前序与中序遍历序列构造二叉树
package com.question.solve.leetcode.programmerCarl2._07_binaryTrees;
import java.util.Deque;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Map;
public class _0105_从前序与中序遍历序列构造二叉树 {
}
/**
* 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 Solution0105 {
Map<Integer, Integer> map;
public TreeNode buildTree(int[] preorder, int[] inorder) {
map = new HashMap<>();
for (int i = 0; i < inorder.length; i++) {//用map保存中序序列的数值对应位置
map.put(inorder[i], i);
}
return findNode(preorder, 0, preorder.length, inorder, 0, inorder.length); //前闭后开
}
public TreeNode findNode(int[] preorder, int preBegin, int preEnd, int[] inorder, int inBegin, int inEnd) {
//参数里的范围都是前闭后开
if (preBegin >= preEnd || inBegin >= inEnd) {//不满足左闭右开，说明没有元素，返回空树
return null;
}
int rootIndex = map.get(preorder[preBegin]); //找到前序遍历的第一个元素在中序遍历中的位置
TreeNode root = new TreeNode(inorder[rootIndex]); //构造结点
int lenOfLeft = rootIndex - inBegin; //保存中序左子树个数，用来确定前序数列的个数
root.left = findNode(preorder, preBegin + 1, preBegin + lenOfLeft + 1,
inorder, inBegin, rootIndex);
root.right = findNode(preorder, preBegin + lenOfLeft + 1, preEnd,
inorder, rootIndex + 1, inEnd);
return root;
}
}
class Solution0105_2 {
public TreeNode buildTree(int[] preorder, int[] inorder) {
if (preorder == null || preorder.length == 0) {
return null;
}
TreeNode root = new TreeNode(preorder[0]);
Deque<TreeNode> stack = new LinkedList<TreeNode>();
stack.push(root);
int inorderIndex = 0;
for (int i = 1; i < preorder.length; i++) {
int preorderVal = preorder[i];
TreeNode node = stack.peek();
if (node.val != inorder[inorderIndex]) {
node.left = new TreeNode(preorderVal);
stack.push(node.left);
} else {
while (!stack.isEmpty() && stack.peek().val == inorder[inorderIndex]) {
node = stack.pop();
inorderIndex++;
}
node.right = new TreeNode(preorderVal);
stack.push(node.right);
}
}
return root;
}
}