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BinaryTreeLevelOrderTraversalII.java
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72 lines (69 loc) · 2.47 KB
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/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
//Use a queue, no recursion, reverse at the last, O(N)
public class Solution {
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
// Start typing your Java solution below
// DO NOT write main() function
LinkedList<TreeNode> nodes = new LinkedList<TreeNode>();
ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();
nodes.add(root);
while(nodes.peek() != null){
LinkedList<TreeNode> tmp = new LinkedList<TreeNode>();
ArrayList<Integer> result = new ArrayList<Integer>();
while(nodes.peek() != null){
TreeNode current = nodes.pop();
if(current.left != null)
tmp.add(current.left);
if(current.right != null)
tmp.add(current.right);
result.add(current.val);
}
results.add(result);
nodes = tmp;
}
ArrayList<ArrayList<Integer>> finalResult = new ArrayList<ArrayList<Integer>>();
for(int i = results.size() - 1; i>=0; i--)
finalResult.add(results.get(i));
return finalResult;
}
}
//Recursion solution, O(N^2)
public class Solution {
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
// Start typing your Java solution below
// DO NOT write main() function
ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();
int max = getMaxLevel(root);
for(int i = max - 1; i >= 0; --i){
ArrayList<Integer> result = new ArrayList<Integer>();
findSameLevel(root, i, result);
results.add(result);
}
return results;
}
public int getMaxLevel(TreeNode root){
if(root!= null)
return Math.max(getMaxLevel(root.left), getMaxLevel(root.right)) + 1;
else
return 0;
}
public void findSameLevel(TreeNode root, int level, ArrayList<Integer> result){
if(root != null){
if(level > 0){
findSameLevel(root.left, level - 1, result);
findSameLevel(root.right, level - 1, result);
}
else{
result.add(root.val);
}
}
}
}