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65 lines (53 loc) · 1.41 KB
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94. Binary Tree Inorder Traversal
Given a binary tree, return the inorder traversal of its nodes values.
For example:
Given binary tree [1,null,2,3],
1
\
2
/
3
return [1,3,2].
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; }
* }
*/
// recursive
public class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
traverse(root, res);
return res;
}
public void traverse(TreeNode node, ArrayList<Integer> res) {
if(node!=null) {
traverse(node.left, res);
res.add(node.val);
traverse(node.right, res);
}
}
}
/////////////////////////////////////////////////////////////////
public class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();
Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode cur = root;
while(cur!=null || !stack.empty()){
while(cur!=null){
stack.add(cur);
cur = cur.left;
}
cur = stack.pop();
list.add(cur.val);
cur = cur.right;
}
return list;
}
}