LeetCode 563. Binary Tree Tilt

Given the root of a binary tree, return the sum of every tree node's tilt.

The tilt of a tree node is the absolute difference between the sum of all left subtree node values and all right subtree node values. If a node does not have a left child, then the sum of the left subtree node values is treated as 0. The rule is similar if the node does not have a right child.

Example 1:

Input: root = [1,2,3]
Output: 1
Explanation: 
Tilt of node 2 : |0-0| = 0 (no children)
Tilt of node 3 : |0-0| = 0 (no children)
Tilt of node 1 : |2-3| = 1 (left subtree is just left child, so sum is 2; right subtree is just right child, so sum is 3)
Sum of every tilt : 0 + 0 + 1 = 1

Example 2:

Input: root = [4,2,9,3,5,null,7]
Output: 15
Explanation: 
Tilt of node 3 : |0-0| = 0 (no children)
Tilt of node 5 : |0-0| = 0 (no children)
Tilt of node 7 : |0-0| = 0 (no children)
Tilt of node 2 : |3-5| = 2 (left subtree is just left child, so sum is 3; right subtree is just right child, so sum is 5)
Tilt of node 9 : |0-7| = 7 (no left child, so sum is 0; right subtree is just right child, so sum is 7)
Tilt of node 4 : |(3+5+2)-(9+7)| = |10-16| = 6 (left subtree values are 3, 5, and 2, which sums to 10; right subtree values are 9 and 7, which sums to 16)
Sum of every tilt : 0 + 0 + 0 + 2 + 7 + 6 = 15

Example 3:

Input: root = [21,7,14,1,1,2,2,3,3]
Output: 9

Constraints:

• The number of nodes in the tree is in the range [0, 104].
• -1000 <= Node.val <= 1000

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这道题是求一棵树中每个节点的左右子树所有节点之和的差值之和……大概就是要求每个节点的左子树的sum和右子树的sum，计算这俩的diff，然后计算所有节点的diff之和。

所以最直观的方法就是写一个recursive sum函数，计算以一个节点为root的这棵树的所有节点value之和。然后再写一个recursive求tilt的函数用来更新总值。代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int result = 0;
    public int findTilt(TreeNode root) {
        tilt(root);
        return result;
    }

    private void tilt(TreeNode root) {
        if (root == null) {
            return;
        }
        int left = sum(root.left);
        int right = sum(root.right);
        int diff = Math.abs(left - right);
        result += diff;
        findTilt(root.left);
        findTilt(root.right);
    }

    private int sum(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return root.val + sum(root.left) + sum(root.right);
    }
}

其实想想就觉得我这个方法递归了两次非常的效率低下，然后看了答案优化了一下。其实可以一边在求sum的时候一边就更新好了result，这其实就相当于post order traversal——先求左子树，再求右子树，再处理根节点。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int result = 0;
    public int findTilt(TreeNode root) {
        sum(root);
        return result;
    }

    private int sum(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int left = sum(root.left);
        int right = sum(root.right);
        int diff = Math.abs(left - right);
        result += diff;
        return root.val + left + right;
    }

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迭代就还是postorder + map处理，很常规，和543很像，感觉终于掌握了postorder迭代的精髓了。

class Solution {
    public int findTilt(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int result = 0;
        Deque stack = new ArrayDeque<>();
        Map map = new HashMap<>();  // node to sum
        stack.push(root);
        // map.put(root, 0);  // seems this is optional
        while (!stack.isEmpty()) {
            TreeNode node = stack.peek();
            if (node.left != null && !map.containsKey(node.left)) {
                stack.push(node.left);
            } else if (node.right != null && !map.containsKey(node.right)) {
                stack.push(node.right);
            } else {
                node = stack.pop();
                map.put(node, map.getOrDefault(node.left, 0) + map.getOrDefault(node.right, 0) + node.val);
                int diff = 0;
                if (node.left != null && node.right != null) {
                    diff = Math.abs(map.get(node.left) - map.get(node.right));
                } else if (node.right != null) {
                    diff = Math.abs(map.get(node.right));
                } else if (node.left != null) {
                    diff = Math.abs(map.get(node.left));
                }
                result += diff;
            }
        }
        return result;
    }
}