Binary Tree

Binary Tree dfs Traversal

-preorder/inorder/postorder

-divide&conquer

-introduce dfs template

Preorder :

Inorder；

Postorder：

Divide & Conquer :

-Merge Sort (arrayO(N)空间, sorted list 不需要) 必会！待补充

-Quick Sort 必会！待补充

-Most of the Binary Tree Problem

题目：

1.Maximum Depth of Binary Tree

2.Balanced Binary Tree

3.Binary Tree Maximum Path Sum

public class Solution {
    int maxPath = Integer.MIN_VALUE;
    public int helper (TreeNode root) {
        if (root == null) {
            return 0;
        }

        int left = helper(root.left);
        int right = helper(root.right);

        maxPath = Math.max(maxPath, Math.max(root.val, Math.max(root.val + left + right, Math.max(root.val + left, root.val + right))));
        return Math.max(root.val, Math.max(root.val + left, root.val + right));//第一遍写成return maxPath;?
    }

    public int maxPathSum(TreeNode root) {
        helper(root);
        return maxPath;
    }
}

黄老师不用全局变量版本:

public class Solution {
    public class ResultType {
        int singlePath, maxPath;
        ResultType(int singlePath, int maxPath) {
            this.singlePath = singlePath;
            this.maxPath = maxPath;
        }
    }

    public ResultType helper (TreeNode root) {
        if (root == null) {
            return new ResultType(0, Integer.MIN_VALUE);
        }

        ResultType left = helper(root.left);
        ResultType right = helper(root.right);

        int single = Math.max(left.singlePath, right.singlePath) + root.val;
        single = Math.max(single, 0);

        int max = Math.max(left.maxPath, right.maxPath);
        max = Math.max(max, left.singlePath + right.singlePath + root.val);

        return new ResultType(single, max);
    }

    public int maxPathSum(TreeNode root) {
        ResultType res = helper(root);
        return res.maxPath;
    }
}

DFS Template：

Template 1: Traverse

public class Solution {
    public void traverse(TreeNode root) {
        if (root == null) {
            return;
        }
        // do something with root
        traverse(root.left);
        // do something with root
        traverse(root.right);
        // do something with root
    }
}

Tempate 2: Divide & Conquer

public class Solution {
    public ResultType traversal(TreeNode root) {
        // null or leaf
        if (root == null) {
            // do something and return;
        }

        // Divide
        ResultType left = traversal(root.left);
        ResultType right = traversal(root.right);

        // Conquer
        ResultType result = Merge from left and right.
        return result;
    }
}

4.Binary  Tree Level Order Traversal

public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();

        if (root == null) {
            return res;
        }

        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.add(root);

        while (!q.isEmpty()) {
            ArrayList<Integer> level = new ArrayList<Integer>();
            int size = q.size();
            for (int i = 0; i < size; i++) {
                TreeNode r = q.poll();
                level.add(r.val);
                if (r.left != null) {
                    q.add(r.left);
                }
                if (r.right != null) {
                    q.add(r.right);
                }
            }
            res.add(level);
        }
        return res;
    }
}