二叉树相关算法：二

LeetCode 437.：Path Sum III

第一个想法就是遍历二叉树，找出每个节点的可能性。这个思路时间复杂度为O(n^2)。

	private int num;
    public int pathSum(TreeNode root, int sum) {
        if(root == null) return num;
        recur(root, sum, 0);
        pathSum(root.left, sum);
        pathSum(root.right, sum);
        return num;
    }
    private void recur(TreeNode root, int sum, int add) {
        if(root == null) {
            return;
        }
        add += root.val;
        if (add == sum) num++;
        recur(root.left, sum, add);
        recur(root.right, sum, add);
    }

改进下代码

	public int pathSum2(TreeNode root, int sum) {
        if(root == null) return 0;
        return recur(root, sum) + pathSum2(root.left, sum) + pathSum2(root.right, sum);
    }

    private int recur(TreeNode node, int sum) {
        if (node == null) return 0;
        return (node.val == sum ? 1: 0) + recur(node.left, sum-node.val) +
                recur(node.right, sum - node.val);
    }

考虑用额外空间来缩短时间复杂度，采用map，其key为从根节点到该节点这条路径上所有节点值的和，value记录该值在整个树中一共有几条这样的路径（都是从根节点开始）。则递归过程中符合要救的路径的个数:
count=map.get(sum-target) ? 0 : 1;

	public int pathSum3(TreeNode root, int target) {
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        map.put(0, 1); //sum-target == 0,则应返回1，表示有一条合格路径
        return recur(root, 0, target, map);
    }
    private int recur(TreeNode node, int sum, int target, Map<Integer, Integer> map){
        if(node == null) return 0;
        sum += node.val;
        // 表示有count条到node的路径(从任意二叉树节点开始)符合要求
        int count = map.getOrDefault(sum-target,0);
        map.put(sum, map.getOrDefault(sum, 0)+1);
        count += recur(node.left, sum, target, map) + recur(node.right, sum, target, map);
        // 回溯过程要清除该节点的记录
        map.put(sum, map.get(sum)-1);
        return count;
    }

538. Convert BST to Greater Tree

    // 解法一：利用中序遍历
    private int sum;

    public TreeNode convertBST(TreeNode root) {
        if(root == null) return null;
        convertBST(root.right);
        sum += root.val;
        root.val = sum;
        convertBST(root.left);
        return root;
    }

解法二

    // 解法二：利用栈

    public TreeNode convertBST2(TreeNode root) {
        int sum = 0;
        ArrayDeque<TreeNode> stack = new ArrayDeque<>();
        TreeNode res = root;
        while (!stack.isEmpty() || root != null) {
            while(root != null) {
                stack.addLast(root);
                root = root.right;
            }

            root = stack.pollLast();
            sum += root.val;
            root.val = sum;
            root = root.left;
        }
        return res;
    }

解法三：

    // 解法三：morris 算法，前面两种解法的时间与空间复杂度均为O(n)
    // morris算法的空间复杂度为O( 1 )
    public TreeNode convertBST3(TreeNode root) {
        int sum = 0;
        TreeNode node = root;
        while (node != null) {
            if (node.right == null) {
                sum += node.val;
                node.val = sum;
                node = node.left;
            }
            else {
                TreeNode succ = getSuccessor(node);
                if (succ.left == null) {
                    succ.left = node;
                    node = node.right;
                }
                else {
                    succ.left = null;
                    sum += node.val;
                    node.val = sum;
                    node = node.left;
                }
            }
        }
        return root;
    }

    private TreeNode getSuccessor(TreeNode node) {
        TreeNode succ = node.right;
        while(succ != null && succ.left != null) {
            succ = succ.left;
        }
        return succ;
    }