樹(2)

• 113. Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Recursive Solution

class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> list = new LinkedList<>();
        List<Integer> path = new LinkedList<>();
        pathSumHelper(list, path, root, sum);
        return list;
    }
    public void pathSumHelper(List<List<Integer>> list, List<Integer> path, TreeNode root, int sum) {
        if (root == null)   return;
        path.add(root.val);
        if (root.left == null && root.right == null) {
            if (root.val == sum)    list.add(new LinkedList(path));
            path.remove(path.size()-1);
            return;
        }
        pathSumHelper(list, path, root.left, sum-root.val);
        pathSumHelper(list, path, root.right, sum-root.val);
        path.remove(path.size()-1);
    }
}

DFS Solution

class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> list = new LinkedList<>();
        if (root == null)   return list;
        List<Integer> path = new LinkedList<>();
        Stack<TreeNode> stack = new Stack<>();
        int sumP = 0;
        TreeNode curNode = root;
        TreeNode visitedNode = null;
        while(curNode != null || !stack.isEmpty()) {
            while (curNode != null) {
                stack.push(curNode);
                sumP += curNode.val;
                path.add(curNode.val);
                curNode = curNode.left;
            }
            curNode = stack.peek();
            if (curNode.right == null || curNode.right == visitedNode) {
                if (curNode.left == null && curNode.right == null && sumP == sum)
                    list.add(new LinkedList(path));
                stack.pop();
                visitedNode = curNode;
                sumP -= curNode.val;
                path.remove(path.size()-1);
                curNode = null;
            } else
                curNode = curNode.right;
        }
        return list;
    }
}

• 235. Lowest Common Ancestor of a Binary Search Tree

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
According to the definition of LCA on Wikipedia: 「The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).」

Recursive Solution

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if ((root.val-p.val)*(root.val-q.val) <= 0) return root;
        if (root.val > p.val)   return lowestCommonAncestor(root.left, p, q);
        return lowestCommonAncestor(root.right, p, q);
    }
}

DFS Solution

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        while((root.val-p.val) * (root.val-q.val) > 0) {
            if (root.val > p.val)   root = root.left;
            else    root = root.right;
        }
        return root;
    }
}

• 404. Sum of Left Leaves

Find the sum of all left leaves in a given binary tree.

BFS Solution

class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        int sum = 0;
        if (root == null || root.left == null && root.right == null)   return sum;
        queue.offer(root);
        while(!queue.isEmpty()) {
            TreeNode node = queue.poll();
            TreeNode left = node.left;
            if (left != null) {
                if (left.left == null && left.right == null) // no need to enqueue the node
                    sum += left.val;
                else
                    queue.offer(node.left);
            }
            if (node.right != null) queue.offer(node.right);
        }
        return sum;
    }
}

Recursive Solution

class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null || root.left == null && root.right == null)    return 0;
        if (root.left != null && root.left.left == null && root.left.right == null)
            return root.left.val + sumOfLeftLeaves(root.right);
        return sumOfLeftLeaves(root.left) + sumOfLeftLeaves(root.right);
    }
}

• 108. Convert Sorted Array to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Recursive Solution

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if(nums == null || nums.length == 0)    return null;
        return getRootHelper(nums, 0, nums.length - 1);
    }
    private TreeNode getRootHelper(int[] nums, int low, int high) {
        if (low > high)    return null;
        int mid = (high-low)/2 + low;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = getRootHelper(nums, low, mid-1);
        root.right = getRootHelper(nums, mid+1, high);
        return root;
    }
}

Iterative: using stack

class Solution {
    private class Node {
        TreeNode node;
        int left, right;
        public Node(TreeNode node, int left, int right) {
            this.node = node;
            this.left = left;
            this.right = right;
        }
    }
    public TreeNode sortedArrayToBST(int[] nums) {
        if (nums == null || nums.length == 0)   return null;
        Stack<Node> stack = new Stack<>();
        TreeNode root = new TreeNode(0);
        Node node = new Node(root, 0, nums.length - 1);
        stack.push(node);
        while (!stack.isEmpty()) {
            Node cur = stack.pop();
            int mid = (cur.right - cur.left) / 2 + cur.left;
            cur.node.val = nums[mid];
            if (mid > cur.left) {
                cur.node.left = new TreeNode(0);
                Node lNode = new Node(cur.node.left, cur.left, mid - 1);
                stack.push(lNode);
            }
            if (mid < cur.right) {
                cur.node.right = new TreeNode(0);
                Node rNode = new Node(cur.node.right, mid + 1, cur.right);
                stack.push(rNode);
            }
        }
        return root;
    }
}

• 543. Diameter of Binary Tree

Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.

分析：一棵樹中最長的路徑有且僅有兩種狀況：通過根節點和不通過根節點。
而通過任意一個節點的最長路徑必定等於左右子樹的最大深度之和。

up to bottom

class Solution {
    public int diameterOfBinaryTree(TreeNode root) {
        if (root == null)   return 0;
        int pathRoot = maxDepth(root.left) + maxDepth(root.right);
        int tmp = Math.max(diameterOfBinaryTree(root.left), diameterOfBinaryTree(root.right));
        return Math.max(tmp, pathRoot);
    }
    private int maxDepth(TreeNode root) {
        if (root == null)   return 0;
        int ldepth = maxDepth(root.left);
        int rdepth = maxDepth(root.right);
        return Math.max(ldepth, rdepth) + 1;
    }
}

可是這種解法中，每call一次都會把當前節點的全部子節點遍歷一遍，有大量重複的計算，時間複雜度爲O(NN)~O(NlgN)（取決於樹的高度）

bottom to up

class Solution {
    private int max = 0;
    public int diameterOfBinaryTree(TreeNode root) {
        maxDepth(root);
        return max;
    }
    private int maxDepth(TreeNode root) {
        if (root == null)   return 0;
        int ldepth = maxDepth(root.left);
        int rdepth = maxDepth(root.right);
        max = Math.max(max, ldepth + rdepth);
        return Math.max(ldepth, rdepth) + 1;
    }
}

這個解法的核心就在於：max = Math.max(max, ldepth + rdepth)這一句，括號中max記錄的是該節點左右兩節點中最大路徑的值，取它和通過該節點的最大路徑中的最大值記錄到max中，因此咱們在調用函數時就不須要再把子節點再遍歷一遍求深度，由於已經記錄在max中了。該算法的時間複雜度爲O(N)。