给定一个二叉树的根节点 root ,返回它的 中序 遍历。
示例 1:
输入:root = [1,null,2,3]
输出:[1,3,2]
示例 2:
输入:root = []
输出:[]
示例 3:
输入:root = [1]
输出:[1]
示例 4:
输入:root = [1,2]
输出:[2,1]
示例 5:
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输入:root = [1,null,2]
输出:[1,2]
提示:
树中节点数目在范围 [0, 100] 内
-100 <= Node.val <= 100
此题主要难点是中序遍历,关于本题解,我采取递归或者迭代的方式完成,并用不同语言实现。
解题思路:从根节点开始,不断向当前节点的左子树遍历并将节点的值加入答案,再递归调用右子树,直到碰到空节点。
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<Integer> inorderTraversal(TreeNode root) {
//定义
List<Integer>list = new ArrayList();
Stack<TreeNode> stack = new Stack<>();
while(root!=null || !stack.isEmpty()){
while(root!=null){
stack.push(root);
root = root.left;
}
root = stack.pop();
list.add(root.val);
root=root.right;
}
return list;
}
}
C语言
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
void inorder(struct TreeNode *root, int * res, int *resSize){
if(!root){
return;
}
inorder(root->left,res,resSize);
res[(*resSize)++] = root->val;
inorder(root->right,res,resSize);
}
int* inorderTraversal(struct TreeNode* root, int* returnSize){
int *res = malloc(sizeof(int)*501);
*returnSize = 0;
inorder(root,res,returnSize);
return res;
}
C++
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> res;
inorder(root,res);
return res;
}
void inorder(TreeNode* root ,vector<int>& res){
if(!root){
return;
}
inorder(root->left,res);
res.push_back(root->val);
inorder(root->right,res);
}
};