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文章目录
- 前言
- [一、力扣971. 翻转二叉树以匹配先序遍历](#一、力扣971. 翻转二叉树以匹配先序遍历)
- [二、力扣987. 二叉树的垂序遍历](#二、力扣987. 二叉树的垂序遍历)
- [三、力扣666. 路径总和 IV](#三、力扣666. 路径总和 IV)
前言
二叉树的递归分为「遍历」和「分解问题」两种思维模式,这道题需要用到「遍历」的思维。
一、力扣971. 翻转二叉树以匹配先序遍历
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<Integer> res = new ArrayList<>();
int[] voyageArr;
int i = 0;
boolean flag = true;
public List<Integer> flipMatchVoyage(TreeNode root, int[] voyage) {
voyageArr = voyage;
fun(root);
if(flag == false){
List<Integer> list = new ArrayList<>();
list.add(-1);
return list;
}
return res;
}
public void fun(TreeNode root){
if(root == null){
return;
}
if(root.val != voyageArr[i++]){
flag = false;
return;
}
if(root.left != null && root.left.val != voyageArr[i]){
res.add(root.val);
TreeNode temp = root.left;
root.left = root.right;
root.right = temp;
}
fun(root.left);
fun(root.right);
}
}
二、力扣987. 二叉树的垂序遍历
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 {
class Triple{
TreeNode node;
int row, col;
public Triple(TreeNode node, int row, int col){
this.node = node;
this.row = row;
this.col = col;
}
}
LinkedList<List<Integer>> res = new LinkedList<>();
List<Triple> nodes = new ArrayList<>();
public List<List<Integer>> verticalTraversal(TreeNode root) {
traverse(root,0,0);
Collections.sort(nodes, (tri1,tri2)->{
if(tri1.col != tri2.col){
return tri1.col - tri2.col;
}else if(tri1.row != tri2.row){
return tri1.row - tri2.row;
}else{
return tri1.node.val - tri2.node.val;
}
});
int pre = Integer.MIN_VALUE;
for(int i = 0; i < nodes.size(); i ++){
Triple cur = nodes.get(i);
if(cur.col != pre){
res.addLast(new LinkedList<Integer>());
pre = cur.col;
}
res.getLast().add(cur.node.val);
}
return res;
}
public void traverse(TreeNode root, int row, int col){
if(root == null){
return ;
}
nodes.add(new Triple(root,row,col));
traverse(root.left, row+1, col-1);
traverse(root.right, row+1, col + 1);
}
}
三、力扣666. 路径总和 IV
java
class Solution {
Map<Integer,Integer> tree = new HashMap<>();
int sum = 0;
public int pathSum(int[] nums) {
for(int i = 0; i < nums.length; i ++){
int value = nums[i]%10;
int code = nums[i]/10;
tree.put(code,value);
}
int rootCode = nums[0]/10;
fun(rootCode,0);
return sum;
}
public void fun(int code, int path){
if(!tree.containsKey(code)){
return;
}
int value = tree.get(code);
int[] pos = decode(code);
int depth = pos[0], id = pos[1];
int leftCode = encode(depth+1, 2*id-1);
int rightCode = encode(depth+1,2*id);
if(!tree.containsKey(leftCode) && !tree.containsKey(rightCode)){
sum += path + value;
return;
}
fun(leftCode, path + value);
fun(rightCode, path + value);
}
public int[] decode(int code){
int id = code%10;
int depth = code/10;
return new int[]{depth,id};
}
public int encode(int depth, int id){
return depth*10 + id;
}
}