Problem
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]
Solution
/**
* 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 result = new Array();
let rooot = new Array();
rooot = root;
if(root){
console.log(rooot.val);
if(root.left == null && root.right == null){
result.push(root.val);
}
else if(root.left != null || root.right != null){
result = inordertraversal(root,result);
}
}
return result;
};
function inordertraversal(root,arr){
if(!root){
return null;
}
else{
inordertraversal(root.left,arr);
arr.push(root.val);
inordertraversal(root.right,arr);
}
return arr;
}
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