LeetCode之二叉树作题java
100. Same Tree
Given two binary trees, write a function to check if they are equal or not.
Two binary trees are considered equal if they are structurally identical and the nodes have the same value.
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
//递归调用,首先判断是否都是空,如果是,则返回true,否则若都不为空,则分两步
//走,如果根节点值相同,则看两者的左节点是否一样,再看右节点,若有不同,则返回false;
//还有就是其他情况:两树中有一树为空的情况,直接返回false
public class Solution {
public boolean isSameTree(TreeNode root1, TreeNode root2){
if(root1==null&&root2==null)
return true;
while(root1!=null&&root2!=null){
if(root1.val==root2.val){
return isSameTree(root1.left, root2.left)&&isSameTree(root1.right, root2.right);
}else{
return false;
}
}
return false;
}
}
101. Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following is not:
1
/ \
2 2
\ \
3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
minimum-depth-of-binary-tree
题目描述
思路:与求二叉树的深度类似,求二叉树的深度,主要是求二叉树的最长路径,此处求二叉树的最短路径,思路如下:
1.当root=null时,直接return 0;
2.当root.left=null且root.right=null时,直接return 1;
3.当root.left=null或者root.right=null时,递归调用(返回右子树最小长度)run(root.right)+1或(返回左子树最小长度)run(root.left)+1;
4.最后root.left!=null且root.right!=null时,返回左子树和右子树的最小路径长度;
public class Solution {
public int run(TreeNode root) {
if(root==null)
return 0;
if(root.left==null&&root.right==null)
return 1;
if(root.left==null)
return run(root.right)+1;
if(root.right==null)
return run(root.left)+1;
return Math.min(run(root.left),run(root.right))+1;
}
}
Given a binary tree, return the postorder traversal of its nodes' values.
For example:
Given binary tree{1,#,2,3},
1
\
2
/
3
return[3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?
考察二叉树的后序遍历:
非递归思路:使用栈来作为辅助
思路一:stack作为缓冲,保存左右孩子节点,stack1作为最终保存结果;
1.当root为空,直接返回空;
2.当root不为空时,当root入栈stack,直接出栈stack,如果该节点没有左右孩子,直接将该节点入栈stack1,最后出栈;
3.如果该节点有左右孩子,将其左右孩子依次入栈stack,保存访问过的节点,重复上面的过程,保证每次取栈顶的元素,左孩子在右孩子前面被访问,左孩子和右孩子都在根节点前面被访问。
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import java.util.*;
public class Solution {
public ArrayList<Integer> postorderTraversal(TreeNode root) {
ArrayList<Integer> list = new ArrayList<Integer>();
if(root==null)
return list;
Stack<TreeNode> stack = new Stack<TreeNode>();
Stack<TreeNode> stack2 = new Stack<TreeNode>();
stack.push(root);
while(!stack.empty()){
TreeNode curNode = stack.pop();
if(curNode.left!=null){
stack.push(curNode.left);
}
if(curNode.right!=null){
stack.push(curNode.right);
}
stack2.push(curNode);
}
while(!stack2.empty()){
list.add(stack2.pop().val);
}
return list;
} }
思路二:对于任一结点P,将其入栈,然后沿其左子树一直往下搜索,直到搜索到没有左孩子的结点,此时该结点出现在栈顶,但是此时不能将其出栈并访问,因此其右孩子还为被访问。所以接下来按照相同的规则对其右子树进行相同的处理,当访问完其右孩子时,该结点又出现在栈顶,此时可以将其出栈并访问。这样就保证了正确的访问顺序。可以看出,在这个过程中,每个结点都两次出现在栈顶,只有在第二次出现在栈顶时,才能访问它。因此需要多设置一个变量标识该结点是否是第一次出现在栈顶。
递归思路:
1.递归遍历左子树;
2.递归遍历右子树;
3.输出根节点;
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import java.util.*;
public class Solution {
public ArrayList<Integer> postorderTraversal(TreeNode root) {
ArrayList<Integer> list = new ArrayList<Integer>();
postorderTraversalHelper(root,list);
return list;
}
public void postorderTraversalHelper(TreeNode root, ArrayList<Integer> list){
if(root==null)
return;
postorderTraversalHelper(root.left,list);
postorderTraversalHelper(root.right,list);
list.add(root.val);
}
}
Given a binary tree, return the preorder traversal of its nodes' values.
For example:
Given binary tree{1,#,2,3},
1
\
2
/
3
return[1,2,3].
Note: Recursive solution is trivial, could you do it iteratively?
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import java.util.*; public class Solution {
public ArrayList<Integer> preorderTraversal(TreeNode root) {
ArrayList<Integer> list = new ArrayList<Integer>();
preorderTraversalHelper(root, list);
return list;
}
public void preorderTraversalHelper(TreeNode root, ArrayList<Integer> list){
if(root==null)
return;
list.add(root.val);
preorderTraversalHelper(root.left,list);
preorderTraversalHelper(root.right,list);
}
}
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