513. 找树左下角的值

给定一个二叉树的 根节点 root，请找出该二叉树的 最底层 最左边 节点的值。

假设二叉树中至少有一个节点。

示例 1:

输入: root = [2,1,3]
输出: 1

/**
 * 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 max = Integer.MIN_VALUE;
    int result = 0;
    private void helper(TreeNode root, int depth) {
        if (root.left == null && root.right == null) {
            if (depth > max) {
                max = depth;
                result = root.val;
            }
            return;
        }

        if (root.left != null) {
            depth++;
            helper(root.left, depth);
            depth--;
        }
        if (root.right != null) {
            depth++;
            helper(root.right, depth);
            depth--;
        }
    }
    public int findBottomLeftValue(TreeNode root) {
        helper(root, 0);
        return result;
    }
}

112. 路径总和

给你二叉树的根节点 root 和一个表示目标和的整数 targetSum 。判断该树中是否存在 根节点到叶子节点 的路径，这条路径上所有节点值相加等于目标和 targetSum 。如果存在，返回 true ；否则，返回 false 。

叶子节点 是指没有子节点的节点。

/**
 * 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 {
    public boolean hasPathSum(TreeNode root, int targetSum) {
        if (root == null) return false;
        targetSum -= root.val;
        if (root.left == null && root.right == null) {
            return targetSum == 0;
        }
        if (root.left != null) {
            boolean left = hasPathSum(root.left, targetSum);
            if (left) {
                return true;
            }
        }
        if (root.right != null) {
            boolean right = hasPathSum(root.right, targetSum);
            if (right) {
                return true;
            }
        }
        return false;
    }
}

113. 路径总和ii

力扣题目链接(opens new window)

给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。

说明: 叶子节点是指没有子节点的节点。

示例: 给定如下二叉树，以及目标和 sum = 22，

/**
 * 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 {
    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) return res; // 非空判断

        List<Integer> path = new LinkedList<>();
        preorderdfs(root, targetSum, res, path);
        return res;
    }

    public void preorderdfs(TreeNode root, int targetsum, List<List<Integer>> res, List<Integer> path) {
            path.add(root.val);

            if (root.left == null && root.right == null) {
            // 找到了和为 targetsum 的路径
                if (targetsum - root.val == 0) {
                    res.add(new ArrayList<>(path));
                }
                return; // 如果和不为 targetsum，返回
            }

            if (root.left != null) {
               preorderdfs(root.left, targetsum - root.val, res, path);
               path.remove(path.size() - 1);
            }
             if (root.right != null) {
            preorderdfs(root.right, targetsum - root.val, res, path);
            path.remove(path.size() - 1); // 回溯
        }
    }
    
}

105
给定两个整数数组 inorder 和 postorder ，其中 inorder 是二叉树的中序遍历， postorder 是同一棵树的后序遍历，请你构造并返回这颗 二叉树 。

class Solution {
    Map<Integer, Integer> map;
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        map = new HashMap<>();
       for (int i = 0; i < inorder.length; i++) {
           map.put(inorder[i], i);
       }
       return findNode(inorder,  0, inorder.length, postorder,0, postorder.length);
    }

    private TreeNode findNode(int[] inorder, int inBegin, int inEnd, int[] postorder, int postBegin, int postEnd) {
        if (inBegin >= inEnd || postBegin >= postEnd) {  // 不满足左闭右开，说明没有元素，返回空树
            return null;
        }
         int rootIndex = map.get(postorder[postEnd - 1]); 
           TreeNode root = new TreeNode(inorder[rootIndex]);  // 构造结点
        int lenOfLeft = rootIndex - inBegin;  // 保存中序左子树个数，用来确定后序数列的个数
        root.left = findNode(inorder, inBegin, rootIndex,
                            postorder, postBegin, postBegin + lenOfLeft);
        root.right = findNode(inorder, rootIndex + 1, inEnd,
                            postorder, postBegin + lenOfLeft, postEnd - 1);

        return root;

    }
}