[Algorithm] 94. Binary Tree Inorder Traversal iteratively approach

Given a binary tree, return the inorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,3,2]

Follow up: Recursive solution is trivial, could you do it iteratively?

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */

function getLefts(root) {
    if (!root) {
        return [];
    }

    let result = [];
    let current = root;
    while(current) {
        result.push(current);
        current = current.left ;
    }

    return result;
}

/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var inorderTraversal = function(root) {
    let lefts = getLefts(root);
    let result = [];
    while (lefts.length) {
        let newRoot = lefts.pop();
        result.push(newRoot.val);
        if (newRoot.right) {
            let newLefts = getLefts(newRoot.right);
            lefts = lefts.concat(newLefts)
        }
    }

    return result;
};

The idea is

• get all the left side node and push into stack
• because stack is first in last out, when we do pop(), we are actually start from bottom of the tree.
• everytime, we pop one node from stack, we check its right child, get all the left nodes from this right child and append into stack.
• in this way, we are able to track all the node.