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binary-tree-preorder-traversal.js
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/**
* Source: https://leetcode.com/problems/binary-tree-preorder-traversal/
* Tags: [Tree,Stack]
* Level: Medium
* Title: Binary Tree Preorder Traversal
* Auther: @imcoddy
* Content: Given a binary tree, return the preorder traversal of its nodes' values.
*
*
* For example:
* Given binary tree {1,#,2,3},
*
* 1
* \
* 2
* /
* 3
*
*
*
* return [1,2,3].
*
*
* Note: Recursive solution is trivial, could you do it iteratively?
*/
/**
* Definition for binary tree
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @returns {number[]}
*/
/**
* Memo: Iterative Solution
* Runtime: 139ms
* Rank: A
*/
var preorderTraversal2 = function(root) {
if (!root) {
return [];
}
var stack = [];
var result = [];
stack.push(root);
while (stack.length) {
var node = stack.pop();
result.push(node.val);
if (node.right) {
stack.push(node.right);
}
if (node.left) {
stack.push(node.left);
}
}
return result;
};
/**
* Memo: Recursive Solution
* Runtime: 141ms
* Rank: A
*/
var preorderTraversal = function(root) {
function traverse(root, result) {
if (!root) {
return;
}
result.push(root.val);
traverse(root.left, result);
traverse(root.right, result);
}
var result = [];
traverse(root, result);
return result;
};
/**
* Memo: Recursive solution
* Complex: O(n)
* Runtime: 136ms
* Tests: 67 test cases passed
* Rank: A
* Updated: 2015-06-06
*/
var preorderTraversal = function(root) {
function traverse(root) {
if (!root) return;
result.push(root.val);
traverse(root.left);
traverse(root.right);
}
var result = [];
traverse(root, result);
return result;
};
function TreeNode(val) {
this.val = val;
this.left = this.right = null;
}
var node = new TreeNode(1);
var root = node;
node = new TreeNode(2);
root.right = node;
node = new TreeNode(3);
root.right.left = node;
console.log(preorderTraversal(root));