Given the root of a binary tree, return the inorder traversal of its nodes’ values.
Example 1:
Input: root = [1,null,2,3] Output: [1,3,2]
Example 2:
Input: root = [] Output: []
Example 3:
Input: root = [1] Output: [1]
Example 4:
Input: root = [1,2] Output: [2,1]
Example 5:
Input: root = [1,null,2] Output: [1,2]
Constraints:
- The number of nodes in the tree is in the range [0, 100].
- -100 <= Node.val <= 100
Follow up: Recursive solution is trivial, could you do it iteratively?
Runtime: 68 ms, faster than 96.16% of JavaScript online submissions for Binary Tree Inorder Traversal.
Memory Usage: 38.7 MB, less than 55.06% of JavaScript online submissions for Binary Tree Inorder Traversal.
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var inorderTraversal = function(root) {
let node = root;
const stack = [];
const result = [];
while(node || stack.length > 0) {
while(node) {
stack.push(node);
node = node.left;
}
if (stack.length > 0) {
node = stack.pop();
result.push(node.val);
node = node.right;
}
}
return result;
};