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?
题意:不使用递归的方式,中序遍历二叉树
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
let inorderTraversal = (root) => {
let res = [];
if (root == null) return res;
let stack = [];
var node = root;
while (node != null || stack.length != 0) {
while (node != null) {
stack.push(node);
node = node.left;
}
if (stack.length != 0) {
node = stack.pop();
res.push(node.val)
node = node.right;
}
}
return res;
}